Shoshan 1
Fundamentals of Quantum Mechanics
Malkiel Shoshan
Shoshan 2
Contents
1. Introduction .............................................................................................................................................. 3
2. Mathematics of Vector Spaces ................................................................................................................. 4
2.1 Linear Vector Spaces ........................................................................................................................... 4
2.2 Inner Product Spaces .......................................................................................................................... 5
2.3 Linear Operators ................................................................................................................................. 6
2.4 Vector Components ............................................................................................................................ 7
2.5 Matrix Representation: Vectors and Operators ................................................................................. 7
2.6 Matrix Representation of the Inner Product and Dual Spaces ........................................................... 9
2.7 The Adjoint Operation: Kets, Scalars, and Operators ....................................................................... 10
2.8 The Projection Operator and the Completeness Relation ................................................................ 10
The Kronecker Delta and the Completeness Relation ........................................................................ 12
2.9 Operators: Eigenvectors and Eigenvalues ........................................................................................ 13
2.10 Hermitian Operators ....................................................................................................................... 14
2.11 Functions as Vectors in Infinite Dimensions ................................................................................... 16
2.12. Orthonormality of Continuous Basis Vectors and the Dirac 𝛅 Function........................................ 17
2.13. The Dirac 𝛅 and the Completeness Relation.................................................................................. 19
2.14. The Dirac 𝛅 Function: How it looks ................................................................................................ 19
The Derivative of the 𝛅 Function ........................................................................................................ 19
The δ Function and Fourier Transforms.............................................................................................. 21
2.15. The Operators K and X ................................................................................................................... 21
2.16. The propogator .............................................................................................................................. 26
3. The Postulates of Quantum Mechanics .................................................................................................. 30
3.1. The Postualtes .................................................................................................................................. 30
3.2. The continuous Eigenvalue Spectrum.............................................................................................. 32
3.3 Collapse of the Wave Function and the Dirac Delta Function .......................................................... 32
3.4. Measuring X and P ........................................................................................................................... 32
3.5. Schrodinger’s Equation .................................................................................................................... 35
4. Particle in a Box ....................................................................................................................................... 38
Conclusion ................................................................................................................................................... 39
Bibliography ................................................................................................................................................ 40
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1. Introduction
The highly confident and optimistic attitude during the era before the quantum revolution was
that all known phenomena were explained by physics. The physicist Albert Michelson, in the 1800s said:
“…The more important fundamental laws and facts of physical science have all been discovered and
these are now so firmly established that the possibility of their ever being supplanted in consequence of
new discoveries is exceedingly remote.” And then, quantum mechanics came and shook the world of
physics at its very foundations.
Many different experiments were designed that demonstrated the inadequacy of classical
physics. They will not be enumerated here; we discuss only one. Most of the mysterious and peculiar
features are contained in the double slit experiment. We first do the (thought) experiment with bullets.
1
P1
2
P2
We have a source (gun), and two walls. Assume the walls are very, very far apart. The bullets
shoot off in different directions. Some go through the slits and hit the next wall. We close slit 2, and
measure the frequency of hits; we get a curve, P1. The curve has a maximum opposite slit 1 that is
where the bullets are most frequent. If we open both slits we get the curve P12.
1
P1
2
P2
Next, we repeat the experiment with light. By light, P1 and P2 interfere, so we get the wavy
pattern. The bullets, which are “lumps” (as Feynman liked to call them), don’t interfere.
If we stop here, everything will be fine. But, we repeat the experiment a third time, this time
with electrons. If we open only one slit we get P1 or P2 as expected. However, if we open both slits we
get the wavy pattern we got by light! How is this possible? Let’s look at the point represented by the red
dot in the figure. When only slit 2 was open, that point got many electrons. How is it that when another
slit is open that point gets fewer electrons? According to classical physics, each electron has a definite
trajectory. The electron going through slit 2 should follow that trajectory; they should not care if slit 1 is
open. Worse still, they should not even know that slit 1 is open. We will see that every particle has a
wave function associated with it that contains all the information about the particle.
These are the conundrums that perplexed physicists in the early 1900s. Many scientists
contributed to the development of quantum mechanics; Albert Einstein, Erwin Schrodinger, Paul Dirac,
Werner Heisenberg and Richard Feynman to name just a few.(At least) Three different but equivalent
mathematical treatments were developed. We use a combination of two approaches, Schrodinger’s
Wave Mechanics and Heisenberg’s Matrix Mechanics. First, we lay the mathematical foundation. Then,
the postulates of quantum mechanics will be explained, and a short example will be presented.
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2. Mathematics of Vector Spaces
2.1 Linear Vector Spaces
Definition: A linear vector space, 𝕍, is a collection of objects called vectors.
A vector, V, is denoted by |𝑉⟩ , read ket V, and satisfies the following properties:
1. They have a rule for addition: |𝑉⟩ + |𝑊⟩
2. They have a rule for scalar multiplication: a|𝑉⟩
3. (closure) The result of these operations is an element of the same vector space 𝕍.
4. (distributive property in vectors) a(|𝑉⟩ + |𝑊⟩) = a|𝑉⟩ + 𝑎|𝑊⟩
5. (distributive property in scalars) (a + b)|𝑉⟩ = 𝑎|𝑉⟩ + 𝑏|𝑉⟩
6. (commutative) |𝑉⟩ + |𝑊⟩ = |𝑊⟩ + |𝑉⟩
7. (associative) |𝑉⟩ + (|𝑊⟩ + |𝑈⟩) = (|𝑉⟩ + |𝑊⟩) + |𝑈⟩
8. A null vector |0⟩ exists so that |𝑉⟩ + |0⟩ = |𝑉⟩
9. An additive inverse vector exists for every |𝑉⟩ so that |𝑉⟩ + |−𝑉⟩ = |0⟩
The familiar arrows from physics obey the above axioms; they are definitely vectors, but not of
the most general form: vectors need not have a magnitude or direction. In fact, matrices also satisfy the
above properties, and, thus, they are also legitimate vectors.
Definition: If we have a set of vectors |𝑉1 ⟩ , |𝑉2 ⟩, … , |𝑉𝑛 ⟩, that are related linearly by
𝑛
�𝑎𝑖 |𝑉𝑖 ⟩ = |0⟩ ,
𝑖=1
(2.1.1)
and (2.1.1) is true only if all the scalars ai = 0, then the set of vectors is linearly independent.
Consider two parallel vectors, |1⟩ and |2⟩, in the xy-plane. One can be written as a multiple of
the other; accordingly, they are not linearly independent. If they were not parallel, they would be
linearly independent. If we now bring in a third vector, |3⟩, into the plane, |3⟩ = 𝑎|1⟩ + 𝑏|2⟩ for some
scalars a and b; thus, |1⟩, |2⟩, and |3⟩ do not form a linearly independent set.
Definition: The dimension, n, of a vector space is the maximum number of linearly independent
vectors it can contain.
Definition: A basis is a set of n linearly independent vectors in an n – dimensional space.
Theorem 1. Any vector, |𝑉⟩, in n – dimensional space can be written linearly in terms of a basis
in that space so that if |1⟩, |2⟩, … , |𝑛⟩ constitute a basis and the vi are scalars, then
𝑛
|𝑉⟩ = � 𝑣𝑖 |𝑖⟩
𝑖=1
(2.1.2)
Proof. If |𝑉⟩ can’t be written in terms of a basis, the n – dimensional space will contain
n+1
linearly independent vectors; the n vectors of a basis and the vector |𝑉⟩ itself. However, by definition an
n – dimesional space can only accommodate n linearly independent vectors.
Definition: The coefficients, vi, in (2.1.2) are called the components of |𝑉⟩ in the |𝑖⟩ basis.
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Definition: A subset of a vector space, 𝕍, that forms a vector space is called a subspace.
A subspace 𝕍 of dimension n is written as 𝕍n. We may give a subspace a label such as 𝕍1𝑦 . The
subscript 𝑦 is the label; it tells us that 𝕍1𝑦 contains all vectors along the 𝑦 axis. The superscript tells us
that the subspace is one dimensional. 𝕍1𝑦 is a subspace of 𝕍3(R) which represents 3 – d space.
2.2 Inner Product Spaces
Recall, we have an operation between two arrows in the plane called the dot product, denoted
by 𝑎⃗ ∙ 𝑐⃗ , that produces a number and satisfies the following three properties:
•
•
•
�⃗ = 𝐵
�⃗ ∙ 𝐴⃗ (symetry)
𝐴⃗ ⋅ 𝐵
𝐴⃗ ∙ 𝐴⃗ ≥ 0
�⃗ + 𝑐𝐶⃗� = 𝑏𝐴⃗ ∙ 𝐵
�⃗ + 𝑐𝐴⃗ ∙ 𝐶⃗ (linearity)
𝐴⃗ ∙ �𝑏𝐵
We now define an operation called the inner product between any two vectors |𝑉⟩ and |𝑊⟩. It is
denoted by ⟨𝑉|𝑊⟩, and satisfies three properties that are analogous to those of the dot product:
•
•
•
⟨𝑉|𝑊⟩ = ⟨𝑊|𝑉⟩* (skew symetry; the * says: take the complex conjugate)
⟨𝑉|𝑉⟩ ≥ 0
⟨𝑉|(𝑎|𝑊⟩ + 𝑏|𝑍⟩) = 𝑎⟨𝑉|𝑊⟩ + 𝑏⟨𝑉|𝑍⟩ (linearity)
The first axiom says the inner product depends on the order of the vectors; the results are
complex conjugates of each other. As with arrows where 𝑎⃗ ∙ 𝑎⃗ is the norm or length squared of the
vector, we want ⟨𝑉|𝑉⟩ to also represent the length squared of the vector, so ⟨𝑉|𝑉⟩ better be positive;
the second axiom expresses this. In addition, ⟨𝑉|𝑉⟩ should be real. The first axiom ensures this:
⟨𝑉|𝑉⟩ = ⟨𝑉|𝑉⟩*
𝑎 + 𝑏𝑖 = 𝑎 − 𝑏𝑖
where i = √−1, and a, b are constants. (2.2.1) can only be true if b = 0 so that ⟨𝑉|𝑉⟩ = 𝑎 is real.
(2.2.1)
The third axiom expresses the linearity of the inner product when a linear combination of
vectors occurs in the second factor. The axiom says that the scalars a, b,…, can come out of the inner
product. If a linear combination occurs in the first factor, we first envoke skew-symetry:
⟨𝑎𝑊 + 𝑏𝑍|𝑉⟩ = ⟨𝑉|𝑎𝑊 + 𝑏𝑍⟩*
and then we use the third axiom
⟨𝑉|𝑎𝑊 + 𝑏𝑍⟩* = (𝑎⟨𝑉|𝑊⟩ + 𝑏⟨𝑉|𝑍⟩)* = 𝑎*⟨𝑉|𝑊⟩* + 𝑏*⟨𝑉|𝑊⟩*
= 𝑎*⟨𝑊|𝑉⟩ + 𝑏*⟨𝑊|𝑉⟩
(2.2.2)
In short, when there is a linear combination in the first factor, the scalars in the first factor get
complex conjugated and taken out of the inner product.
Definition: Two vectors are orthogonal if their inner product is zero.
Definition: The norm of a vector is expressed as �⟨𝑉|𝑉⟩ ≡ |𝑉|
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Definition: A set of basis vectors, all of unit norm and mutually orthogonal, is referred to as an
orthonormal basis.
We can now derive a formula for the inner product. Given |𝑉⟩ and |𝑊⟩ with components vi and
wj, respectively, written in terms of an orthonormal basis |1⟩, |2⟩, … , |𝑛⟩,
|𝑉⟩ = � 𝑣𝑖 |𝑖⟩
𝑖
|𝑊⟩ = � 𝑤𝑗 | 𝑗⟩,
𝑗
⟨𝑉|𝑊⟩ = �� 𝑣𝑖 |𝑖⟩ � � 𝑤𝑗 |𝑗⟩�
𝑖
𝑗
By (2.2.2), the components, vi, of |𝑉⟩ (the vector in the first factor) can come out of the inner
product provided we take their complex conjugate. Thus,
⟨𝑉|𝑊⟩ = ∑𝑖 ∑𝑗 𝑣𝑖 *𝑤𝑗 ⟨𝑖|𝑗⟩
(2.2.3)
However, we still need to know the inner product between the basis vectors, ⟨𝑖|𝑗⟩. We know
that the inner product between an orthonormal vector and itself is, by definition, the norm of the
orthonormal vector which, also by definition, equals one. In addition, the inner product between two
different orthonormal vectors is, by defintion, zero. Thus, the value of ⟨𝑖|𝑗⟩ will be one or zero
depending if i = j or i ≠ j:
⟨𝑖|𝑗⟩ = �
1 if 𝑖 = 𝑗
≡ 𝛿𝑖𝑗
0 if 𝑖 ≠ 𝑗
(2.2.4)
where δij, called the Kronecker delta, is shorthand for the function in (2.2.4). Due to the Kronecker delta,
only the terms that occur with i = j will survive. So, (2.2.3) now collapses to
⟨𝑉|𝑊⟩ = ∑𝑖 𝑣𝑖 * 𝑤𝑖
(2.2.5)
⟨𝑉|𝑉⟩ = ∑𝑖 |𝑣𝑖 |2
(2.2.6)
Accordingly, the inner product between a vector and itself will be
We calculate the norm as defined before, �⟨𝑉|𝑉⟩. A vector is normalized if its norm equals one.
The kronecker delta combines the orthogonality and normalization of orthonormal basis vectors
into one entity. If 𝑖 = 𝑗, we get the inner product between a vector and itself; for the basis vectors this
inner product equals one, so we know they are normalized. If 𝑖 ≠ 𝑗, we get an inner product of zero, so
they are orthogonal. Simply put, if we know that ⟨𝑖|𝑗⟩ = 𝛿𝑖𝑗 , the basis is orthonormal.
2.3 Linear Operators
An operator acts on a vector, |𝑉⟩, and transforms it into another vector |𝑉 ′ ⟩. If Ω is an operator,
we reprsent its action as follows:
Ω|𝑉⟩ = |𝑉 ′ ⟩
(2.3.1)
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We will deal only with linear operators, operators that satisfy the following criterion:
Ω(𝑎|𝑉⟩ + 𝑏|𝑊⟩) = 𝑎Ω|𝑉⟩ + 𝑏Ω|𝑊⟩
(2.3.2)
Two examples of operators include the identity operator, I, and the rotation operator Rπ(k)
I|𝑉⟩ = |𝑉⟩
R|𝒌⟩ = |𝒌⟩
R|𝒊⟩ = −|𝒊⟩
The identity operator leaves the vector unchanged, the rotation operator rotates vectors in 3 – d space
around the z-axis by π.
By (2.1.2), any vector |𝑉⟩ can be written as a linear combination of basis vectors. Therefore if
we know how an operator transforms the basis vectors, we will know how it transform |𝑉⟩. So, if
Ω|𝑖⟩ = |𝑖′⟩
where i is an index for the basis |1⟩, |2⟩, … , |𝑛⟩, then for a vector |𝑉⟩ which can be expressed as
𝑛
|𝑉⟩ = � 𝑣𝑖 |𝑖⟩
𝑖
we have
𝑛
𝑛
𝑛
𝑖=1
𝑖=1
𝑖=1
Ω|V⟩ = Ω � 𝑣𝑖 |𝑖⟩ = � 𝑣𝑖 Ω|𝑖⟩ = � 𝑣𝑖 |𝑖′⟩
2.4 Vector Components
(2.3.4)
To find the components of any vector, |𝑊⟩, in an orthonormal basis |1⟩, |2⟩, … ,|𝑛⟩, we take the
inner product between |𝑊⟩ and the basis vectors:
|𝑊⟩ = �
𝑛
⟨𝑘|𝑊⟩ = �
𝑛
𝑤𝑖 |𝑖⟩
𝑖=1
𝑤𝑖 ⟨𝑘|𝑖⟩ = 𝑤𝑖 𝛿𝑘𝑖
𝑖=1
⟨𝑘|𝑊⟩ = 𝑤𝑖
(2.4.1)
2.5 Matrix Representation: Vectors and Operators
A vector |𝑉⟩ in some basis |1⟩, |2⟩, … , |𝑛⟩, can be represented by an n × 1 column matrix whose
elements are the components of |𝑉⟩ in the given basis. So, if we have 5 basis vectors, and,
then |𝑉⟩ can be represented by
|𝑉⟩ = 5|1⟩ + 3|2⟩ − 4|3⟩ + 2|5⟩
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5
⎡ 3⎤
⎢
⎥
|𝑉⟩ ↔ ⎢ −4 ⎥
⎢ 0⎥
⎣ 2⎦
and the basis vectors, |𝑛⟩, will then be represented by n × 1 column matrices where the nth
element is 1. Then,
1
⎡0⎤
⎢ ⎥
|1⟩ ↔ ⎢0⎥
⎢0⎥
⎣0⎦
and,
|𝑉⟩ ↔
5
⎡ 3⎤
⎢ ⎥
⎢−4⎥
⎢ 0⎥
⎣ 2⎦
0
⎡1⎤
⎢ ⎥
|2⟩ ↔ ⎢0⎥
⎢0⎥
⎣0⎦
1
⎡0⎤
⎢ ⎥
= 5 ⎢0⎥
⎢0⎥
⎣0⎦
0
⎡1⎤
⎢ ⎥
+ 3 ⎢0⎥
⎢0⎥
⎣0⎦
0
⎡0⎤
⎢ ⎥
|5⟩ ↔ ⎢0⎥
⎢0⎥
⎣1⎦
0
⎡0⎤
⎢ ⎥
− 4 ⎢1⎥
⎢0⎥
⎣0⎦
0
⎡0⎤
⎢ ⎥
+ 2 ⎢0⎥
⎢0⎥
⎣1⎦
Just as a vector can be represented by an n × 1 matrix, a linear operator can be represented by
an n × n matrix. If we have a linear operator Ω acting on an orthonormal basis |1⟩ , |2⟩ , … , |𝑛⟩ where
Ω|𝑖⟩ = |𝑖′⟩
then any vector V written in terms of this basis will be transformed, according to (2.3.4), into
𝑛
Ω|𝑉⟩ = � 𝑣𝑖 |𝑖′⟩
𝑖=1
(2.5.1)
According to (2.4.1), to find the components of |𝑖′⟩, the transformed basis vector, in terms of the
original basis vector |𝑖⟩, we take the inner product between them:
⟨𝑗|𝑖′⟩ = ⟨𝑗|Ω|𝑖⟩ ≡ Ω𝑗𝑖
where |𝑗⟩ and |𝑖⟩ are different indices refering to the same basis, and Ω𝑗𝑖 is a shorthand.
We can then build an n × n matrix representing Ω. The Ω𝑗𝑖 from (2.5.2) are the n × n
elements; Ω𝑗𝑖 is the element in the jth column and the 𝑖th row.
If we have
Ω|𝑉⟩ = |𝑉 ′ ⟩,
we can write the new components, vi’ , in terms of the original components, vi , as in (2.5.2):
𝑣𝑖′ = ⟨𝑖|𝑉′⟩ = ⟨𝑖|Ω|𝑉⟩
Since V can be written as a linear combination of its basis vectors, we have
𝑣𝑖′ = ⟨𝑖|Ω|𝑉⟩ = ⟨𝑖|Ω�∑𝑗 𝑣𝑗 |𝑗⟩� = �𝑖� ∑𝑗 𝑣𝑗 Ω �𝑗� = ∑𝑗 𝑣𝑗 ⟨𝑖|Ω|𝑗⟩
(2.5.2)
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and by our shorthand from (2.5.2) we get
𝑣𝑖′ = ∑𝑗 𝑣𝑖 Ω𝑖𝑗
(2.5.3)
According to the rules of matrix multiplication, (2.5.3) is equivalent to the following:
or,
Ω11
𝑣′
⎡ 1⎤
⎡
⎢𝑣2′ ⎥
⎢Ω21
⎢ ⎥ = ⎢
⎢⋮⎥
⎢ ⋮
⎢
⎢ ⎥
⎣Ω𝑛1
⎣𝑣𝑛′ ⎦
Ω12
Ω22
…
…
⋱
…
Ω1𝑛
⎤
⋮ ⎥
⎥
⎥
⎥
Ω𝑛𝑛 ⎦
𝑉′ = Ω𝑉
𝑣1
⎡ ⎤
⎢𝑣2 ⎥
⎢ ⋮ ⎥=
⎢ ⎥
⎣𝑣𝑛 ⎦
⟨1|Ω|1⟩ ⟨1|Ω|2⟩
⎡
⎢⟨2|Ω|1⟩ ⟨2|Ω|2⟩
⎢
⎢ ⋮
⎢
⎣⟨n|Ω|1⟩
…
↔
…
⋱
…
⟨1|Ω|𝑛⟩
⎤
⋮ ⎥
⎥
⎥
⎥
⟨n|Ω|𝑛⟩⎦
|𝑉′⟩ = Ω|𝑉⟩
𝑣1
⎡ ⎤
⎢𝑣2 ⎥
⎢ ⋮ ⎥ (2.5.4)
⎢ ⎥
⎣𝑣𝑛 ⎦
(2.5.5)
These two equations say that the vector, |𝑉′⟩, resulting from the action of Ω on V, can be
represented by the product of the matrix representing Ω and the matrix representing V. Simply put, to
see how an operator, Ω, transforms a vector, V, multiply their matrices.
2.6 Matrix Representation of the Inner Product and Dual Spaces
If we have two vectors |𝑉⟩ and |𝑊⟩ represented by matrices as column vectors
𝑣1
𝑣2
|𝑉⟩ ↔ � ⋮ �
𝑣𝑛
𝑤1
𝑤2
|𝑊⟩ ↔ � ⋮ � ,
𝑤𝑛
we can’t write the inner product, ⟨𝑉|𝑊⟩, as a product of matrices because we can’t multiply two column
vectors together. However, if we take the adjoint or transpose conjugate of the first vector, so that the
matrix for |𝑉⟩ will be
then we can write the inner product as
[𝑣1 * 𝑣2 * ⋯
𝑣𝑛 *]
𝑤1
⎡ ⎤
⎢𝑤2 ⎥
⟨𝑉|𝑊⟩ = [𝑣𝑛 * 𝑣𝑛 * ⋯ 𝑣𝑛 *] ⎢ ⎥
⋮
⎢ ⎥
⎣𝑤𝑛 ⎦
This serves as motivation for the notation attributed to Paul Dirac. Just as we associate with
each column matrix a ket, we will associate with each row matrix a new entity called a bra. Ket V is
written as |𝑉⟩, and bra V is written as ⟨𝑉|. The two entities are adjoints of each other. That is, to get bra
V, take the transpose conjugate of the column matrix representing V as we did above.
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Thus, we have two distinct vector spaces, the space of bras and the space of kets. The inner
product is defined between a bra and a ket. Above we had an inner product between ket |𝑊⟩ and bra
⟨𝑉|; together, ⟨𝑉|𝑊⟩, they form a bra – ket (or bracket).
2.7 The Adjoint Operation: Kets, Scalars, and Operators
If a scalar, a, multiplies a ket |𝑉⟩, so that
𝑎|𝑉⟩ = |𝑎𝑉⟩
then the corresponding bra will be
⟨𝑎𝑉| = ⟨𝑉|𝑎*
We can see why if we look at the matrix representations.
The adjoint of this matrix is
𝑎𝑣1
𝑎𝑣2
𝑎|𝑉⟩ = |𝑎𝑉⟩ ↔ � ⋮ �
𝑎𝑣𝑛
[𝑎*𝑣1 * 𝑎*𝑣2 * ⋯ 𝑎*𝑣𝑛 *] = [𝑣1 * 𝑣2 * ⋯ 𝑣𝑛 *]𝑎* ↔ ⟨𝑉|𝑎*
We now define the adjoint of an operator Ω as follows. If Ω acts on |𝑉⟩ we have
Ω|𝑉⟩ = |Ω𝑉⟩
(2.7.1)
(where |Ω𝑉⟩ is the vector that results after Ω acts on |𝑉⟩). The corresponding bra is
⟨ΩV| = ⟨𝑉|Ω†
(2.7.2)
Ω† is called the adjoint of Ω.
Equations (2.7.1) and (2.7.2) say if Ω acts on ket V to produce |𝑉′⟩, Ω† acts on bra V to produce
|. It follows that Ω† also has a matrix associated with it. We can calculate its elements in a basis as
we did in (2.5.2)
⟨𝑉 ′
By (2.7.2), this becomes
(Ω†)𝑖𝑗 = ⟨𝑖|Ω† |𝑗⟩
(Ω†)𝑖𝑗 = ⟨Ω𝑖|𝑗⟩ = ⟨𝑗|Ω𝑖⟩* = ⟨𝑗|Ω|𝑖⟩* = Ω𝑗𝑖 *
Thus, to find the adjoint of an operator take the transpose conjugate of its matrix.
2.8 The Projection Operator and the Completeness Relation
We know that for any vector |𝑉⟩,
Shoshan 11
𝑛
𝑛
𝑖=1
𝑖=1
|𝑉⟩ = � 𝑣𝑖 |𝑖⟩ = �|𝑖⟩⟨𝑖|𝑉⟩
(2.8.1)
Consider, the object in the summation, |𝑖⟩⟨𝑖|𝑉⟩, gives us the component of V along the 𝑖th basis vector
multiplied by that basis vector. Alternatively, it is the projection of |𝑉⟩ along |𝑖⟩. The entity, |𝑖⟩⟨𝑖|, can be
thought of as the projection operator. It acts on |𝑉⟩ to give
|𝑖⟩⟨𝑖|𝑉⟩ = |𝑖⟩𝑣𝑖
We call |𝑖⟩⟨𝑖| , ℙ𝑖 , so |𝑖⟩⟨𝑖|𝑉⟩ becomes ℙ𝑖 |𝑉⟩. Then we can write (2.8.1) as
𝑛
|𝑉⟩ = � ℙ𝑖 |𝑉⟩.
𝑖=1
In (2.8.2),which operator does (∑𝑛𝑖=1 ℙ𝑖 ) correspond to? If we rewrite (2.8.2) as
𝑛
|𝑉⟩ = �� ℙ𝑖 � |𝑉⟩.
𝑖=1
(2.8.2)
(2.8.3)
We see that after the operator acts on |𝑉⟩, the result remains |𝑉⟩; then the operator must be
the identity operator I. Therefore, we get
𝑛
𝑛
𝑖=1
𝑖=1
𝐼 = � ℙ𝑖 = �|𝑖⟩ ⟨𝑖|
(2.8.4)
Equation (2.8.4) is called the completeness relation. It says the sum of all the projections of a
vector V along its basis vectors is the vector V itself. (2.8.4) expresses the fact that basis vectors are
complete; meaning any vector V can be written as a linear combination of the basis vectors.
This highlights the difference between a basis and any other linearly independent set of vectors.
Any vector can’t necessarily be expressed as a linear combination of a linear independent set, but it
must be able to be expressed as a linear combination of basis vectors.
For example, in 3 – d space, if we have the unit vectors i and j, they form a linearly independent
set but not a basis, and we can’t write the unit vector k in terms of i and j. However, if we have all three
i, j, k they form a basis and any vector can be written as
In short, all basis are complete.
ket
v = ai + bj + ck
We now look at the matrix representation of the operator, ℙ, and of (2.8.4). We have the basis
0
⎡0⎤
⎢ ⎥
⋮
|𝑖⟩ ↔ ⎢ ⎥
1
⎢ ⎥
⎢⋮⎥
⎣0⎦
Shoshan 12
and the basis bra
⟨𝑖| ↔ [0
The projection operator
ℙ = |𝑖⟩⟨𝑖|
is the product of these
0
⎡ ⎤
⎢0⎥
⎢ ⎥
⋮
ℙ = |𝑖⟩⟨𝑖| = ⎢ ⎥ [0
⎢1⎥
⎢ ⎥
⎢⋮⎥
⎣0⎦
0
⎡
⎢0
⎢
⎢⋮
⋯ 0] = ⎢
⎢
⎢
⎢
⎣0
0 ⋯ 1
The sum of all projections will be
1
⎡ ⎤
⎢0⎥
⎢ ⎥
⎢ ⋮ ⎥ [1
⎢0⎥
⎢ ⎥
⎣0⎦
0 … 0 0]
1
⎡
⎢0
= ⎢⋮
⎢
⎣0
0
0
⋯
⋯
⋱
⋯ 0]
0 ⋯ 1
+
0
⎡ ⎤
⎢ 1⎥
⎢ ⎥
⎢0⎥ [0 1
⎢⋮⎥
⎢ ⎥
⎣ 0⎦
0
0
⎤
⎡
⋮⎥
⎢0
⎥ + ⎢⋮
⎥
⎢
⎣0
0⎦
the identity matrix as expected.
0
⋱
0
⎡ ⎤
⎢0⎥
⎢ ⎥
⎢1⎥ [0 0
⎢0⎥
⎢ ⎥
+
0
1
0
0
⎤
⎡
⋮⎥
⎢0
⎥ + ⎢⋮
⎥
⎢
⎣0
0⎦
⋯
1
⎡
⎢0
= ⎢⋮
⎢
⎣0
0
⋱
0
1
⋯
⋯
⋱
1
0
⋱
0
1
⋯
0 ⋯ 0]
⋯
⋯
⎣⋮⎦
0
⎤
⋮⎥
⎥
0⎥
1⎦
0
0
⋯
⋱
1 0
⋯
1
⋱
⎤
⋮⎥
⎥
⎥
⎥
⎥
⎥
⎥
0⎦
⋯]
0
⎤
⋮⎥
⎥
⎥
0⎦
+ ⋯
+ ⋯
The Kronecker Delta and the Completeness Relation
We know that if |𝑥⟩ is a basis vector,
|Ψ⟩ = �|𝑥⟩⟨𝑥|Ψ⟩
𝑥
If |Ψ⟩ itself is a basis vector, |𝑦⟩, the completeness relation should still hold in an orthonormal basis,
Shoshan 13
|y⟩ = �|𝑥⟩⟨𝑥|y⟩
𝑥
But |𝑥⟩ and |𝑦⟩ are basis vectors so ⟨𝑥|𝑦⟩ = 𝛿𝑥𝑦 and
|𝑦⟩ = �|𝑥⟩𝛿𝑥𝑦
𝑥
the δxy will kill all terms except for the term that occurs when x = y , so we’re left with
|𝑦⟩ = |𝑦⟩
Thus, the kronecker delta was defined in a way that is consistent with the completeness relation.
2.9 Operators: Eigenvectors and Eigenvalues
For every operator, Ω, there are vectors that when acted upon by Ω will only undergo a
rescaling and not any other transformation so that
Ω|𝑊⟩ = 𝜔|𝑊⟩
(2.9.1)
The vectors, |𝑊⟩, that satisfy (2.9.1) are called the eigenvectors of Ω, and the scale factors, ω,
are its eigenvalues; (2.9.1) is known as an eigenvalue equation. In general, if |𝑊⟩ is an eigenvector of Ω,
𝑎|𝑊⟩ will also be an eigenvector:
Ω𝑎|𝑊⟩ = 𝑎Ω|𝑊⟩ = 𝑎𝜔|𝑊⟩
Thus, the vector 𝑎|𝑊⟩ has also been rescaled by the factor ω, so, it, too, is an eigenvector. For
now, we look at two examples of operators and their eigenvectors; later we will meet more.
Example 1. The identity operator I
The eigenvalue equation is
𝐼|𝑊⟩ = 𝜔|𝑊⟩
(2.9.2)
We know that if ω = 1, any vector will satisfy (2.9.2) because the vector just remains the same.
Hence, I has an infinite number of eigenvectors; in fact, every vector is an eigenvector of I. I has only
one eigenvalue which is 1; i.e. the scale of the vectors doesn’t change.
Example 2. The rotation operator we mentioned before, Rπ(k)
Once again, this operator rotates all vectos in 3 – d space π radians around the z – axis.
Obviously, the only vectors in real space it doesn’t rotate are the vectors that lay along the z – axis.
Some of its eigenvectors include the unit vector k (or |3⟩ as we have called it before) and any multiple of
k, ak. For these eigenvectors, its eigenvalue is one since it doesn’t rescale.
(2.9.1):
We now develop a method to find solutions to the eigenvalue equation. We first rearrange
(Ω − 𝐼𝜔)|𝑊⟩ = 0
(2.9.3)
Shoshan 14
and
|𝑊⟩ = (Ω − 𝐼𝜔)−1 |0⟩
(2.9.4)
we know that the inverse of a matrix A is
𝐴−1 =
𝐶𝑜𝑓𝑎𝑐𝑡𝑜𝑟 𝐴𝑇
𝐷𝑒𝑡 𝐴
Hence, the only way (2.9.4) can happen is if
𝐷𝑒𝑡(Ω − 𝜔𝐼) = 0
(2.9.5)
After expanding the determinant of (Ω − 𝜔𝐼), we will get a polynomial with powers of ω and
coefficients, a0, a1, …, an; (2.9.5) will be of the form
𝑛
� 𝑎𝑚 𝜔𝑚 = 0
𝑚=0
(2.9.6)
(2.9.6) is known as the characteristic equation. Solving the characteristic equation gives us the
eigenvalues, after which we can find the eigenvectors.
2.10 Hermitian Operators
Definition: An operator, Ω, is hermitian if Ω = Ω†
Because operators can be written as matrices, we also have the following definition.
Definition: A matrix, M, is hermitian if Ω = Ω†
Theorem 2. The eigenvalues of a hermitian operator Ω are real.
Proof. If
Ω|𝑊⟩ = 𝜔|𝑊⟩
we take the inner product of both sides with a bra ⟨𝑊|
⟨𝑊|Ω|𝑊⟩ = 𝜔⟨𝑊|𝑊⟩
and then also take the adjoint of both sides
since Ω is hermitian we have Ω = Ω†
(2.10.1)
⟨𝑊|Ω† |𝑊⟩ = 𝜔*⟨𝑊|𝑊⟩
⟨𝑊|Ω|𝑊⟩ = 𝜔*⟨𝑊|𝑊⟩
We now subtract (2.10.2) from (2.10.1)
⟨𝑊|Ω|𝑊⟩ − ⟨𝑊|Ω|𝑊⟩ = 𝜔⟨𝑊|𝑊⟩ − 𝜔*⟨𝑊|𝑊⟩
(2.10.2)
Shoshan 15
we are left with
0 = (𝜔 − 𝜔*)⟨𝑊|𝑊⟩
ω = ω*
the only way this is possible is if ω is real.
Theorem 3. The eigenvectors of every hermitian operator form an orthonormal basis; in this
basis, Ω, can be written as a diagonal matrix with its eigenvalues as diagonal entries.
Proof: Following the method of solving eigenvalue equations developed earlier, we solve the
characteristic equation of Ω. We will get at least one root ω1 that corresponds to an eigenvector |𝑊1 ⟩.
Assuming Ω operates in the vector space 𝕍n, we look at the subspace 𝕍n-1 that is orthogonal to |𝑊1 ⟩.
We then take any n – 1 orthonormal vectors from 𝕍n-1 and the original normalized |𝑊1 ⟩, and form a
basis.
In terms of this basis, Ω looks like
𝜔1
⎡
⎢0
⎢
Ω ↔ ⎢0
⎢ ⋮
⎢
0
0
⋯
⎣0
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(2.10.3)
The first element is ω1 because (as was discussed in section 2.5 on operators)
Ω11 = ⟨1|Ω|1⟩,
here the first basis vector is |𝑊1 ⟩, so we have
Ω11 = ⟨𝑊1 |Ω|𝑊1 ⟩ = 𝜔1 ⟨𝑊1 |𝑊1 ⟩ = 𝜔1
The rest of the column is zero because the basis is orthogonal; thus, if |𝑖⟩ is any of the other basis
vectors,
⟨𝑖|Ω|𝑊1 ⟩ = 𝜔1 ⟨𝑖|𝑊1 ⟩ = 0
The rest of the row is zero because Ω is hermitian and must be symetric along the diagonal. The blank
part of the matrix consists of elements about to be determined.
Consider, the blank part of the matrix in (2.10.3), which we’ll call M, can itself be a matrix of
dimension n – 1; we can solve its eigenvalue earlier too. Once again, the characteristic equation yields a
root, ω2, corresponding to an eigenvector (which we normalize) |𝑊2 ⟩. Once again, we choose a
subspace one dimension less, 𝕍n-2, that is orthogonal to |𝑊2 ⟩. We choose any n – 2 orthonormal vectors
from 𝕍n-2 and our original |𝑊2 ⟩ to form a basis.
In terms of this basis, M looks like
Shoshan 16
𝜔2
⎡
⎢0
⎢
𝑀 = ⎢0
⎢ ⋮
⎢
Then, the original operator, Ω, looks like
0
0
⋯
0
⎣0
𝜔
⎡ 1
⎢0
⎢
Ω ↔ ⎢ ⋮
⎢
⎢
⎣0
0
𝜔2
0
⋯
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(2.10.4)
0
⎤
0⎥
⎥
⎥
⎥
⎥
⎦
⋯
⋮
0
We then repeat the process on the blank part in Ω until we are finally left with
Ω ↔
𝜔
⎡ 1
⎢0
⎢
⎢0
⎢
⎢ ⋮
⎢
⎢
⎣0
0
𝜔2
0
⋮
0
0
⋯
0
⋯
0
⋱
0
…
𝜔3
⋮
0
0
⋯
𝜔𝑛−1
0
⎤
0⎥
⎥
0⎥
⎥
⋮ ⎥
0⎥
⎥
𝜔𝑛 ⎦
The eigenvectors |𝜔1 ⟩, |𝜔2 ⟩, … , |𝜔𝑛 ⟩ form an orthonormal basis because each subspace 𝕍n-1,
𝕍 , … , 𝕍1 was chosen to be orthogonal to the previous one.
n-2
2.11 Functions as Vectors in Infinite Dimensions
Going back to our basic properties of vectors, we notice that there are functions that satisfy
these properties. If we have two functions, f(x) and g(x), there is a definite rule for addition, there is a
rule for scalar multiplication, we design functions that have inverses, and, in short follow all the
properties enumerated in section 2.1.
We first look at discrete functions. Let us look at a function f. It is defined at x1, x2, … , xn with
values f(x1), f(x2), … , f(xn), respectively. The basis vectors, |𝑥⟩, will constitute of functions that are
nonzero at only one point, xi with the value f(xi) = 1.
The basis vectors are orthonormal that is,
and obey the completeness relation
⟨𝑥𝑖 |𝑥𝑗 ⟩ = 𝛿𝑖𝑗
�|𝑥𝑖 ⟩⟨𝑥𝑖 | = 𝐼
𝑖
(2.11.1)
Shoshan 17
To find the component of f along a basis vector |𝑥⟩, we follow the typical method, we take the inner
product between f and that basis vector
⟨𝑥|𝑓⟩ = 𝑓(𝑥).
We can, as in (2.8.1), write f as a linear combination of basis vectors
|𝑓⟩ = �|𝑥𝑖 ⟩⟨𝑥𝑖 |𝑓⟩
𝑖
= � 𝑓(𝑥𝑖 ) |𝑥𝑖 ⟩
𝑖
The inner product between functions will be the same as before
⟨𝑓|𝑔⟩ = � 𝑓*(𝑥𝑖 )𝑔(𝑥𝑖 )
𝑖
(2.11.2)
(2.11.3)
Now, we would like to extend the above to continuous functions. The basis vectors are the same
as before only now there is a basis vector for every point x, i.e. there is an infinite number of basis
vectors. A function f will then be a vector in infinite dimensions. Continuous functions have relations
similar to equations (2.11.1), (2.11.2), (2.11.3); just in each case the summation is replaced by an
integral.
The completeness relation will then be
�|𝑥⟩⟨𝑥|𝑑𝑥 = 𝐼
(2.11.2) will become
|𝑓⟩ = �𝑓(𝑥)|𝑥⟩ 𝑑𝑥
and the inner product will be
⟨𝑓|𝑔⟩ = � 𝑓*(𝑥)𝑔(𝑥)𝑑𝑥
2.12. Orthonormality of Continuous Basis Vectors and the Dirac 𝛅 Function
Basis vectors are orthogonal, so
⟨𝑥|𝑦⟩ = 0
What is the normalization of the basis vectors? Should ⟨𝑥|𝑦⟩ = 1 ? To answer this question we start with
the completeness relation.
𝑏
�|𝑦⟩⟨𝑦|𝑑𝑦 = 𝐼
𝑎
We then take the inner product with a ket |𝑓⟩ and a bra ⟨𝑥|
Shoshan 18
𝑏
�⟨𝑥|𝑦⟩⟨𝑦|𝑓⟩𝑑𝑦 = ⟨𝑥|𝑓⟩
𝑎
𝑏
�⟨𝑥|𝑦⟩𝑓(𝑦)𝑑𝑦 = 𝑓(𝑥)
𝑎
Let us call ⟨𝑥|𝑦⟩ an unknown function 𝛿(𝑥, 𝑦)
𝑏
� 𝛿(𝑥, 𝑦)𝑓(𝑦)𝑑𝑦 = 𝑓(𝑥)
𝑎
We know that 𝛿(𝑥, 𝑦) is zero wherever 𝑥 ≠ 𝑦 (since they’re orthogonal), so we can limit the integral to
the infinitesimal region where 𝑥 = 𝑦
𝑥+𝛼
� 𝛿(𝑥, 𝑦)𝑓(𝑦)𝑑𝑦 = 𝑓(𝑥)
𝑥−𝛼
We see that the integral of the delta function together with 𝑓(𝑦) is 𝑓(𝑥). If we take the limit as
𝑓(𝑦) → 𝑓(𝑥). Then, in the limit, we can pull 𝑓(𝑦) out of the integral as 𝑓(𝑥)
(2.12.1)
α → 0,
𝑥+𝛼
𝑓(𝑥) � 𝛿(𝑥, 𝑦)𝑑𝑦 = 𝑓(𝑥)
𝑥−𝛼
𝑥+𝛼
(2.12.2)
� 𝛿(𝑥, 𝑦)𝑑𝑦 = 1
𝑥−𝛼
So 𝛿(𝑥, 𝑦) is a very peculiar function. It’s zero everywhere except for one point, 𝑥 = 𝑦. At that point
it can’t be finite because then the integral in (2.12.2), which is over an infinitesimal region, will be zero
not one. At 𝑥 = 𝑦, 𝛿(𝑥, 𝑦) must be infinite in such a way that the integral sums to one.
Nowhere did we care about the actual values of 𝑥 and 𝑦, only about whether they were equal or
not. This leads us to denote the delta function by 𝛿(𝑥 − 𝑦). It’s known as the Dirac Delta function with
the following values.
𝒃
𝛿(𝑥 − 𝑦) = 0
� 𝛿(𝑥 − 𝑦)dy = 1
𝒂
when
𝑥≠𝑦
(2.12.3𝑎)
when 𝑥 = 𝑦
(2.12.3𝑏)
Thus, ⟨𝑥|𝑦⟩ = 𝛿(𝑥 − 𝑦). In the continuous case, basis vectors are not normalized to one, rather
they are normalized to the Dirac delta function.
Shoshan 19
2.13. The Dirac 𝛅 and the Completeness Relation
If |𝑥⟩ and |𝑦⟩ are basis vectors, we have
𝑏
|𝑦⟩ = �|𝑥⟩⟨𝑥|𝑦⟩𝑑𝑥
𝑎
𝑏
|𝑦⟩ = �|𝑥⟩𝛿(𝑥 − 𝑦)𝑑𝑥
𝑎
(2.13.1)
If ⟨𝑥|𝑥⟩ would equal one, as in the discrete case, that is, if ⟨𝑥|𝑥⟩ = 𝛿(𝑥 − 𝑥) = 𝛿(0) = 1, (2.13.1)
would not be true. However, now that 𝛿(0) does not equal one, rather it follows (2.12.3b), (2.13.1) is of
the form of the following equation shown earlier:
𝑥+𝛼
� 𝛿(𝑥, 𝑦)𝑓(𝑦)𝑑𝑦 = 𝑓(𝑥)
𝑥−𝛼
(2.12.1)
The Dirac delta collapses the integral to the one term when 𝑥 = 𝑦. It is the counterpart of the Kronecker
delta. The Kronecker delta collapses sums; the Dirac delta collapses integrals.
2.14. The Dirac 𝛅 Function: How it looks
From, the discussion above it should be clear that 𝛿(𝑥 − 𝑥′) [where |𝑥⟩ and |𝑥′⟩ are basis
vectors] is sharply peaked at 𝑥 and zero elsewhere. The sharp peak comes from the fact that the area
under it must total one (2.12.3b).
It helps to think of the deltafunction as a limit of the Gaussian function:
Taking the limit
We have the familiar bell curve of width Δ. The area under it is one. The delta function could
then be the limit of the Gaussian as Δ → 0.
The area would still be one, and we’ll have a very sharp peak at 𝑥. The peak has to be high “enough” so
that the area should still be one. The delta function has a value of zero or infinite.
The Derivative of the 𝛅 Function
We examine the derivative of 𝛿(𝑥 − 𝑥′) because it will be used later. The derivative is denoted
by
Shoshan 20
𝑑
𝛿(𝑥 − 𝑥′) = 𝛿′(𝑥 − 𝑥′)
𝑑𝑥
We first look at its graph, and as before we compare it to the Gaussian. The derivative of the
Gaussian, graph A, will be
𝑑𝑔(𝑥 ′ − 𝑥)
𝑑𝑥
𝑓
When we take the limit of it as Δ → 0, we get the delta derivative in graph B
𝑥
Graph A is the Delta function spiked at 𝑥. In B there are two rescaled δ functions, one is positive
at 𝑥 − 𝛼, one is negative at 𝑥 + 𝛼 .
We had earlier (2.12.1), when we integrate the delta function with another function we get back
the function itself. In particular for a coordinate 𝑥
𝑏
� 𝛿(𝑥 − 𝑥 ′ )𝑓(𝑥 ′ )𝑑𝑥′ = 𝑓(𝑥)
𝑎
the δ function picks out the value 𝑓(𝑥). We see this in Graph C.
The derivative of δ, 𝛿 ′ (𝑥 − 𝑥 ′ ), will pick out two values, one at 𝑥 − 𝛼 and one at 𝑥 + 𝛼; one
gives the value 𝑓(𝑥 − 𝛼), one gives the value – 𝑓(𝑥 + 𝛼)
We then have
� 𝛿′(𝑥 − 𝑥 ′ )𝑓(𝑥 ′ )𝑑𝑥 ′ =
The numerator will be equal to
𝑓(𝑥 − 𝛼) − 𝑓(𝑥 + 𝛼)
2𝛼
𝑓(𝑥 − 𝛼) − 𝑓(𝑥 + 𝛼) = 2𝛼
𝑑𝑓(𝑥)
𝑑𝑥 ′
(2.14.1)
(2.14.2)
This is the amount the function 𝑓 changes by over the interval [𝑥 − 𝛼, 𝑥 + 𝛼]. Putting (2.14.1)
into (2.14.2) we get
� 𝛿′(𝑥 − 𝑥 ′ )𝑓(𝑥 ′ )𝑑𝑥 ′ =
𝑑𝑓(𝑥)
𝑑𝑥 ′
(2.14.3)
For comparison, the integral of 𝛿(𝑥 − 𝑥 ′ ) with𝑓(𝑥 ′ ) gives us 𝑓(𝑥), the integral of 𝛿′(𝑥 − 𝑥′)
with 𝑓(𝑥 ′ ) gives us 𝑓′(𝑥). Thus,
Shoshan 21
𝛿 ′ (𝑥 − 𝑥 ′ )𝑓(𝑥 ′ )
gives the same result as
𝛿(𝑥 − 𝑥 ′ )
because
𝑑
𝑓(𝑥 ′ )
𝑑𝑥 ′
� 𝛿(𝑥 − 𝑥 ′ )
𝑑
𝑓(𝑥 ′ )
𝑑𝑥 ′
= � 𝛿(𝑥 − 𝑥 ′ )𝑓 ′ (𝑥 ′ ) = 𝑓′(𝑥)
so we can write
𝛿 ′ (𝑥 − 𝑥 ′ ) = 𝛿(𝑥 − 𝑥 ′ )
𝑑
𝑑𝑥
The δ Function and Fourier Transforms
From Fourier analysis we have Fourier’s Inversion Theorem that says
∞
∞
−∞
−∞
1
� 𝑒 𝑖𝜔𝑡 𝑑𝜔 � 𝑓(𝑢)𝑒 −𝑖𝜔𝑢 𝑑𝑢
𝑓(𝑡) =
2𝜋
∞
∞
1
= � 𝑓(𝑢)𝑑𝑢 �
� 𝑒 𝑖𝜔(𝑡−𝑢) 𝑑𝜔�
2𝜋
−∞
−∞
Comparing this to (2.12.1) we see that
∞
1
� 𝑒 𝑖𝜔(𝑡−𝑢) 𝑑𝜔
𝛿(𝑡 − 𝑢) =
2𝜋
−∞
2.15. The Operators K and X
If a function 𝑓 is a vector, we can have an operator Ω that operates on 𝑓 to produce another
function
Ω|𝑓⟩ = |𝑓̃⟩
Later on, we will look at two important operators, 𝑋 and 𝐾. For now, we look at the
differentiation operator 𝐷.
𝐷|𝑓⟩ = �
𝑑𝑓
� = |𝑓′⟩
𝑑𝑥
(2.15.1)
We know that we can compute the matrix elements, Ω𝑖𝑗 , of an operator Ω by taking the inner
product with basis kets and bras.
Ω𝑖𝑗 = ⟨𝑖|Ω|𝑗⟩
Therefore, we can compute the matrix elements 𝐷𝑥𝑥′ of 𝐷 by
Shoshan 22
𝐷𝑥𝑥' = ⟨𝑥|𝐷|𝑥′⟩
We start with (2.15.1), and take the inner product with a basis bra ⟨𝑥|,
⟨𝑥|𝐷|𝑓⟩ =
𝑑𝑓(𝑥)
𝑑𝑥
We then write the left side in terms of the completeness relation
�|𝑥′⟩⟨𝑥 ′ |𝑑𝑥 ′ ⟨𝑥|𝐷|𝑓⟩ = �⟨𝑥|𝐷|𝑥 ′ ⟩⟨𝑥′|𝑓⟩𝑑𝑥 ′ =
= �⟨𝑥|𝐷|𝑥 ′ ⟩𝑓(𝑥 ′)𝑑𝑥′ =
𝑑𝑓(𝑥)
𝑑𝑥
𝑑𝑓(𝑥)
𝑑𝑥
(2.15.2)
This equation (2.15.2) is very similar to something we had earlier
If we compare, we see that
or
� 𝛿′(𝑥 − 𝑥 ′ )𝑓(𝑥 ′ )𝑑𝑥 ′ =
𝑑𝑓(𝑥)
𝑑𝑥 ′
(2.14.3)
⟨𝑥|𝐷|𝑥′⟩ = 𝛿′(𝑥 − 𝑥 ′ )
𝐷𝑥𝑥' = 𝛿(𝑥 − 𝑥 ′ )
𝑑
𝑑𝑥'
𝐷, then, is a matrix of infinite dimensions. Is 𝐷 Hermitian? If it were, we would have
Let us check, we know
and
But since the δ function is real
∗
𝐷𝑥𝑥' = 𝐷𝑥'𝑥
𝐷𝑥𝑥 ' = 𝛿′(𝑥 − 𝑥 ′ )
(2.15.3)
∗
= 𝛿′(𝑥′ − 𝑥)*
𝐷𝑥'𝑥
and since it’s an odd function
𝛿′(𝑥′ − 𝑥)* = 𝛿′(𝑥′ − 𝑥)
so
𝛿′(𝑥′ − 𝑥) = − 𝛿′(𝑥 − 𝑥′)
∗
= − 𝛿′(𝑥 − 𝑥′)
𝐷𝑥'𝑥
Comparing equations (2.15.3) and (2.15.4) we see that
∗
𝐷𝑥𝑥' = − 𝐷𝑥'𝑥
(2.15.4)
Shoshan 23
Therefore, 𝐷 is not Hermitian. However, we can make 𝐷 Hermitian by multiplying it by −𝑖. We
call the resulting operator 𝐾,
𝐾 = −𝑖𝐷 = −𝑖𝛿′(𝑥 − 𝑥′)
For 𝐾, we have
but, once again, δ is real so
∗
∗
𝐾𝑥'𝑥
= −𝑖𝐷𝑥'𝑥
= −𝑖𝛿′(𝑥′ − 𝑥)*
and since it’s odd
∗
= 𝑖𝛿′(𝑥′ − 𝑥)
𝐾𝑥'𝑥
∗
𝐾𝑥'𝑥
= −𝑖𝛿′(𝑥 − 𝑥′)
but the right side is what we had for K, then
∗
𝐾𝑥𝑥' = 𝐾𝑥'𝑥
As of now it seems that 𝐾 is Hermitian, but we now use a different approach. If we have two
vectors |𝑓⟩ and |𝑔⟩, we should have
by skew symetry we have
⟨𝑔|𝐾|𝑓⟩ = ⟨𝑔|𝐾𝑓⟩
from section 2.7 we know that
⟨𝑔|𝐾𝑓⟩ = ⟨𝐾𝑓|𝑔⟩*
(2.15.5)
⟨𝐾𝑓| = ⟨𝑓|𝐾 †
and
But, if 𝐾 is Hermitian,
⟨𝐾𝑓|𝑔⟩* = ⟨𝑓|𝐾 † |𝑔⟩*
⟨𝑓|𝐾 † |𝑔⟩* = ⟨𝑓|𝐾|𝑔⟩*
(2.15.6)
Taking equations (2.15.5) and (2.15.6), we conclude that if 𝐾 is Hermitian, we must have
⟨𝑔|𝐾|𝑓⟩ = ⟨𝑓|𝐾|𝑔⟩*
We rewrite both sides in terms of inegrals: we interpose the completeness relation twice, so we
get double integrals. Is the following equation true?
𝑏
𝑏
∗
�⟨𝑔|𝑥⟩⟨𝑥|𝐾|𝑥′⟩⟨𝑥′|𝑓⟩𝑑𝑥 ′ 𝑑𝑥 = ��⟨𝑓|𝑥⟩⟨𝑥|𝐾|𝑥 ′ ⟩⟨𝑥 ′ |𝑔⟩𝑑𝑥 ′ 𝑑𝑥 �
𝑎
𝑎
We now simplify the integrals. We start with the left side. First, by skew symetry
⟨𝑔|𝑥⟩ = ⟨𝑥|𝑔⟩*
(2.15.7)
Shoshan 24
this gives the component of |𝑔⟩ along |𝑥⟩
⟨𝑥|𝑔⟩* = 𝑔*(𝑥)
also,
⟨𝑥 ′ |𝑓⟩ = 𝑓(𝑥 ′ ).
Next, we have
⟨𝑥|𝐾|𝑥′⟩ = 𝐾𝑥𝑥' = −𝑖𝐷𝑥𝑥' = −𝑖𝛿 ′ (𝑥 − 𝑥 ′ ) = −𝑖𝛿(𝑥 − 𝑥 ′ )
𝑑
𝑑𝑥'
this operates on whatever follows it in the integral, i.e. 𝑓(𝑥 ′ ). Then the inner integral on the left
side of (2.15.7) looks like
𝑏
� 𝑔*(𝑥)(−𝑖)𝛿(𝑥 − 𝑥 ′ )
𝑎
𝑏
𝑑
𝑓(𝑥 ′ )𝑑𝑥 ′
𝑑𝑥'
−𝑖𝑑𝑓(𝑥)
= 𝑔*(𝑥)(−𝑖) � 𝛿(𝑥 − 𝑥 ′ )𝑓 ′ (𝑥 ′ )𝑑𝑥 ′ = 𝑔*(𝑥) �
�
𝑑𝑥
𝑎
Once again, the delta function kills off the integral. For the outer integral we have
𝑏
−𝑖𝑑𝑓(𝑥)
� 𝑑𝑥
� 𝑔*(𝑥) �
𝑑𝑥
𝑎
We do the same simplification to the right side of (2.15.7) and we’re left with
𝑏
𝑏
−𝑖𝑑𝑔(𝑥)
−𝑖𝑑𝑓(𝑥)
� 𝑔*(𝑥) �
� 𝑑𝑥 = �� 𝑓*(𝑥) �
� 𝑑𝑥 �
𝑑𝑥
𝑑𝑥
𝑎
Due the conjugation, the right side becomes
𝑏
𝑖 � 𝑓(𝑥)
𝑎
∗
𝑎
𝑑𝑔*(𝑥)
𝑑𝑥
𝑑𝑥
We now integrate the left side of (2.15.8) by parts with the following substitutions
𝑢 = −𝑖𝑔*(𝑥)
𝑑𝑢 = −𝑖
𝑑𝑔*(𝑥)
𝑑𝑥
𝑑𝑣 =
𝑑𝑓(𝑥)
𝑑𝑥
𝑑𝑥
𝑣 = 𝑓(𝑥)
(2.15.8)
Shoshan 25
Finally, we have
[−𝑖𝑔*(𝑥)𝑓(𝑥)]𝑏𝑎
or
𝑏
𝑏
𝑑𝑔*(𝑥)
𝑑𝑔*(𝑥)
+ 𝑖 � 𝑓(𝑥)
𝑑𝑥 = 𝑖 � 𝑓(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑎
𝑎
[−𝑖𝑔*(𝑥)𝑓(𝑥)]𝑏𝑎 = 0
(2.15.9)
If this equation is true, 𝐾 is Hermitian, meaning 𝐾 is Hermitian in a space of functions that
satisfy this equation.There are two sets of functions that obey (2.15.9): functions that are zero at
endpoints, and periodic functions that follow
𝑓(𝜃) = 𝑓(𝜃 + 2𝜋)
For the first type, we have
[−𝑖𝑔*(𝑥)𝑓(𝑥)]𝑏𝑎 = 0 − 0 = 0
For the second type, we have
[−𝑖𝑔*(𝑥)𝑓(𝑥)]2𝜋
0 = −𝑖𝑔*(2𝜋)𝑓(2𝜋) + 𝑖𝑔*(0)𝑓(0) = 0
We now find the eigenvectors of 𝐾.
𝐾|Ψ𝑘 ⟩ = 𝑘|Ψ𝑘 ⟩
To find components in the |𝑥⟩ basis, we follow the typical procedure: taking the inner product
with a bra ⟨𝑥|
⟨𝑥|𝐾|Ψ𝑘 ⟩ = 𝑘⟨𝑥|Ψ𝑘 ⟩
�⟨𝑥|𝐾|𝑥′⟩⟨𝑥′|Ψ𝑘 ⟩ = 𝑘⟨𝑥|Ψ𝑘 ⟩
−𝑖
𝑑
Ψ (𝑥) = 𝑘Ψ𝑘 (𝑥)
𝑑𝑥 𝑘
We know that the function whose derivative is the function itself is the exponential
Ψ𝑘 (𝑥) = 𝐴𝑒 𝑖𝑘𝑥 =
1
√2𝜋
𝑒 𝑖𝑘𝑥
Thus, 𝐾 has an infinite number of eigenvectors |𝑘⟩, one for every real number. We choose 𝐴 to
normalize them to the Dirac δ
∞
⟨𝑘|𝑘′⟩ = �
−∞
⟨𝑘|𝑥⟩⟨𝑥|𝑘 ′⟩𝑑𝑥
2
∞
= 𝐴 � 𝑒 −𝑖�𝑘−𝑘
−∞
′ �𝑥
𝑑𝑥 =
1
𝛿(𝑘 − 𝑘 ′ )
2𝜋
We now look at the 𝑋 operator. Its action on a vector is to multiply the vector by 𝑥, so
𝑋|𝑓⟩ = 𝑥|𝑓⟩
Shoshan 26
By dotting both sides with a basis bra ⟨𝑥|, we have, in the |𝑥⟩ basis
⟨𝑥|𝑋|𝑓⟩ = 𝑥⟨𝑥|𝑓⟩ = 𝑥𝑓(𝑥)
(2.15.10)
We now find its eigenvectors. The eigenvalue equation is
𝑋|𝑥⟩ = 𝑥|𝑥⟩
In the |𝑥⟩ basis
⟨𝑥′|𝑋|𝑥⟩ = 𝑥⟨𝑥′|𝑥⟩ = 𝑥𝛿(𝑥 ′ − 𝑥)
Comparing this to (2.15.10) we see that the eigenvectors are
𝑋 is a Hermitian operator because
1
2.16. The propogator
|𝑥⟩ = 𝛿(𝑥 ′ − 𝑥)
1
� 𝑓*(𝑥)𝑥𝑔(𝑥)𝑑𝑥 = �[𝑥𝑓*(𝑥)]𝑔(𝑥)𝑑𝑥
−1
−1
We now explain and demonstrate the concept of a propogator through the use of an example.
Suppose we have a vibrating string that’s clamped at both ends, 𝑥 = 0, 𝑥 = 𝐿, where 𝐿 is the length of
the string. We would like to know how the string will look in the future.
The function Ψ(𝑥) gives the displacement of the string at the point 𝑥. It can be thought of as a
snapshot of the string at one instant, Ψ(𝑥, 𝑡) will then be a snapshot of the string at time 𝑡. Ψ(𝑥, 𝑡)
changes with time according to
Since 𝐾 = −𝑖
𝜕
𝜕𝑥
𝜕2Ψ
𝜕2Ψ
=
𝜕𝑡 2
𝜕𝑥 2
, the right side can be written as −𝐾 2 |Ψ⟩ so we have
|Ψ̈(𝑡)⟩ = −𝐾 2 |Ψ(𝑡)⟩
(2.16.1)
|Ψ(𝑡)⟩ = 𝑈(𝑡)|Ψ(0)⟩
(2.16.2)
Our goal is to find a function 𝑈(𝑡) such that
Meaning, for any initial state |Ψ(0)⟩ we just multiply it by a function 𝑈(𝑡) and the result will be
|Ψ(𝑡)⟩; 𝑈(𝑡) is called the propogator. The utility of the propogator comes from the fact that once we
find the propogator we have completely solved the problem. The first step in constructing the
propogator is solving the eigenvalue problem of the operator, here −𝐾 2 . The eigenvalue equation is
or
In the |𝑥⟩ basis we have
−𝐾 2 |Ψ⟩ = −𝑘 2 |Ψ⟩
𝐾 2 |Ψ⟩ = 𝑘 2 |Ψ⟩
(2.16.3)
Shoshan 27
𝐾 2 ⟨𝑥|Ψ⟩ = 𝑘 2 ⟨𝑥|Ψ⟩
𝑑2
Ψ (𝑥) = 𝑘 2 Ψ𝑘 (𝑥)
𝑑𝑥 2 𝑘
The solution to this differential equation gives us the eigenvectors
−
Ψ𝑘 (𝑥) = 𝐴 cos 𝑘𝑥 + 𝐵 sin 𝑘𝑥
(2.16.4)
𝐴 and 𝐵 are constants about to be determined. We know that
Ψ𝑘 (0) = 0
then,
Also,
𝐴 cos 𝑘(0) + 𝐵 sin 𝑘(0) = 0
𝐴=0
Ψ𝑘 (𝐿) = 0
So
𝐵 sin 𝑘𝐿 = 0
If we want a solution other than 𝐵 = 0, we must have 𝑘𝐿 equal to a multiple of π.
then (2.16.4) becomes
𝑘𝐿 = 𝑚𝜋
𝑚 = 1, 2, 3, ⋯
𝑚𝜋𝑥
Ψ𝑚 (𝑥) = 𝐵 sin �
�
𝐿
We choose 𝐵 so that Ψ𝑚 (𝑥) is normalized:
𝐿
𝑚𝜋𝑥 2
�� 𝑑𝑥 = 1
� �𝐵 sin �
𝐿
0
𝐿
𝑚𝜋𝑥
� 𝑑𝑥 = 1
𝐵 � 𝑠𝑖𝑛2 �
𝐿
2
𝐿
0
2𝑚𝜋𝑥
𝐵2
� �1 − cos �
�� 𝑑𝑥 = 1
2
𝐿
0
𝐿𝐵2
=1
2
⟹
2
𝐵= �
𝐿
(2.16.5)
Shoshan 28
then (2.16.5) is
𝑚𝜋𝑥
2
Ψ𝑚 (𝑥) = � sin �
�
𝐿
𝐿
We now have an eigenbasis made of the eigenvectors
𝑚𝜋 2
𝑚𝜋𝑥
2
�
|𝑚⟩ → � sin �
𝐿
𝐿
in the |𝑥⟩ basis with eigenvalues � � . The eigenbasis is orthonormal, i.e. ⟨𝑚|𝑛⟩ = 𝛿𝑚𝑛 . When 𝑚 = 𝑛,
𝑙
⟨𝑚|𝑛⟩ = 1, because we normalized the vectors above. if 𝑚 ≠ 𝑛, we have
𝐿
Making the substitution 𝑢 =
𝜋𝑥
𝐿
using the trigonometric identity
𝑚𝜋𝑥
𝑛𝜋𝑥
2
⟨𝑚|𝑛⟩ = � sin �
� sin �
� 𝑑𝑥
𝐿
𝐿
𝐿
0
𝜋
2
⟨𝑚|𝑛⟩ = � sin(𝑚𝑢) sin(𝑛𝑢) 𝑑𝑢
𝜋
0
sin 𝐴 sin 𝐵 =
𝜋
1
[cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)]
2
𝜋
1
⟨𝑚|𝑛⟩ = �� cos(𝑚𝑢 − 𝑛𝑢)𝑑𝑢 − � cos(𝑚𝑢 + 𝑛𝑢)𝑑𝑢�
𝜋
0
=
𝜋
0
𝜋
1 sin(𝑚 − 𝑛)𝑢
1 sin(𝑚 + 𝑛)𝑢
�
� − �
�
𝜋
𝑚− 𝑛
𝜋
𝑚+ 𝑛
0
0
⟨𝑚|𝑛⟩ = 0
Now that we have solved the eigenvalue problem of −𝐾 2 we need to construct the propogator in terms
of the eigenbasis. We write (2.16.1) in this eigenbasis; we dot both sides with bra ⟨𝑚|
𝑑2
⟨𝑚|Ψ(𝑡)⟩ = −𝐾 2 ⟨𝑚|Ψ(𝑡)⟩
𝑑𝑡 2
But by the eigenvalue equation we had before the right side becomes
Shoshan 29
or
𝑚𝜋 2
−𝐾 2 ⟨𝑚|Ψ(𝑡)⟩ = −𝑘 2 ⟨𝑚|Ψ(𝑡)⟩ = − � � ⟨𝑚|Ψ(𝑡)⟩
𝐿
𝑑2
𝑚𝜋 2
⟨𝑚|Ψ(𝑡)⟩
� ⟨𝑚|Ψ(𝑡)⟩
=
−
�
𝑑𝑡 2
𝐿
Solving the differential equation we get
We can now write
𝑚𝜋𝑡
⟨𝑚|Ψ(𝑡)⟩ = ⟨𝑚|Ψ(0)⟩ cos �
�
𝐿
|Ψ(𝑡)⟩ = �|𝑚⟩⟨𝑚|Ψ(𝑡)⟩
𝑚
𝑚𝜋𝑡
�
|Ψ(𝑡)⟩ = �|𝑚⟩⟨𝑚|Ψ(0)⟩ cos �
𝐿
𝑚
comparing this to the propogator equation we had earlier (2.16.2)
we see that
|Ψ(𝑡)⟩ = 𝑈(𝑡)|Ψ(0)⟩
𝑚𝜋𝑡
𝑈(𝑡) = �|𝑚⟩⟨𝑚| cos �
�
𝐿
𝑚
Finally we write the propogator in the |𝑥⟩ basis; we dot both sides with bra ⟨𝑥| and ket |𝑥′⟩
𝑚𝜋𝑡
⟨𝑥|𝑈(𝑡)|𝑥′⟩ = �⟨𝑥|𝑚⟩⟨𝑚|𝑥′⟩ cos �
�
𝐿
𝑚
𝑚𝜋𝑥
𝑚𝜋𝑥′
𝑚𝜋𝑡
2
⟨𝑥|𝑈(𝑡)|𝑥′⟩ = � sin �
� sin �
�
� cos �
𝐿
𝐿
𝐿
𝐿
𝑚
We then go back to (2.16.2) and write it in |𝑥⟩ basis
(2.16.6)
𝐿
⟨𝑥|Ψ(𝑡)⟩ = Ψ(𝑥, 𝑡) = ⟨𝑥|𝑈(𝑡)|Ψ(0)⟩ = �⟨𝑥|𝑈(𝑡)|𝑥′⟩⟨𝑥′|Ψ(0)⟩ 𝑑𝑥 ′ (2.16.7)
0
We substitute what we get for the propogator from (2.16.6) into (2.16.7) and perform the integral.
Shoshan 30
3. The Postulates of Quantum Mechanics
3.1. The Postualtes
The postulates of quantum mechanics are now stated and explained. They are postulates in the
sense that we can’t derive them mathematically. They can be compred to Newton’s third law, 𝐹 = 𝑚𝑎.
Just as Newton’s law can’t be proven, only supported by experiment, these postulates are assumptions
that yield the correct results when tested by experiment.
The postualtes here deal with one particle in one dimension. The first three deal with the state
of the particle at one instant; the fourth deals with how the state evolves with time.
Postulate I. The state of a particle is represented by a vector |Ψ⟩ in the physical Hilbert space.
We know that some vectors are normalizable to one, and some can be normalized to the δ
function. The space of both of these types of vectors is called the physical Hilbert Space.
In classical physics, we just need to know two dynamic variables, 𝑥(𝑡) and 𝑝(𝑡), i.e. position and
momentum, to know everything about the particle. The state of the particle is representes by (𝑥, 𝑝), in 3
�⃗, �𝒑⃗). All information about any dynamic variable, 𝑊(𝑥, 𝑝) can be determined from it. For
– d (𝒓
example, kinetic energy and angular momentum will be
𝑝2
𝐸=
2𝑚
𝑎𝑛𝑑
�𝑳⃗ = 𝒓
�⃗ × 𝒑
�⃗
In quantum mechanics, the state is represnted by what we call a wave a function |Ψ⟩. The wave
function, in the |𝑥⟩ basis, at one instant in time is a function of 𝑥: Ψ(𝑥). This is a vector in infinite
dimensions. Thus, while in classical mechanics we only needed two variables to express the state of the
particle, in quantum mechanics we need an infinite number of them.
We saw how in classical physics we can easily extract information from the state (𝑥, 𝑝). How do
we do the same in quantum mechanics? The second and third posulates tell us just that. The second
postualte defines the equivalence of the classical variables 𝑥 and 𝑝.
Postulate II. The Hermitian operators 𝑋 and 𝑃 represent the classical variables 𝑥 and 𝑝. Other
dynamical variables, 𝑊(𝑥, 𝑝), are represented by Ω(𝑋, 𝑃).
The 𝑋 operator is the same discussed in section 2.15 with matrix elements
The operator 𝑃 is
⟨𝑥|𝑋|𝑥′⟩ = 𝑥𝛿(𝑥 − 𝑥 ′ )
𝑃 = ℏ𝐾 = −𝑖ℏ𝐷
where 𝐾 was discussed in the same section and ℏ = ℎ⁄2𝜋 The matrix elements of 𝑃, in the |𝑥⟩
basis, will then be
⟨𝑥|𝑃|𝑥 ′ ⟩ = −𝑖ℏ𝛿 ′ (𝑥 − 𝑥 ′ )
Postulate III. The measurement of a variable Ω(𝑋, 𝑃), that has eigenvectors |𝜔⟩, will yield an
eigenvalue, ω, of Ω with a probability proportional to |⟨𝜔|𝜓⟩|2 . The state will change from |Ψ⟩ to |𝜔⟩.
Shoshan 31
As was said earlier, postulates II and III tell us how to extract information from the state |Ψ⟩.
First, we follow postulate II to find the operator Ω that corresponds to the dynamic variable we’re
measuring. Then, we solve the eigenvalue problem of Ω, and find all eigenvectors, |𝜔𝑖 ⟩, and
eigenvalues, ωi, of Ω. Since Ω is Hermitian, its eigenvectors form a basis. Then we can expand |Ψ⟩ in
terms of this eigenbasis using the completeness relation. If the eigenvalues are discrete
If they are continuous
|Ψ⟩ = �|𝜔𝑖 ⟩⟨𝜔𝑖 |Ψ⟩
𝑖
|Ψ⟩ = �|𝜔⟩⟨𝜔|Ψ⟩𝑑𝜔
If |Ψ⟩ is normalized, the probability of measuring a value |𝜔⟩ is equal to |⟨𝜔|𝜓⟩|2 , otherwise it is
proportional to it.
We first note the use of two theorems proved earlier. The eigenvalues of Hermititian operators
are real, and the eigenvectors of Hermitian form a basis. If we make a measurement, we expect the
result to be real, and we can always write the state |Ψ⟩ as a linear combination of eigenvectors because
they form a basis.
The quantum theory only allows probablistic predictions, and the only possible values to be
measured are the eigenvalues. We illustrate with an example. Suppose we have an operator Ω and a
state |Ψ⟩ written in terms of the eigenvectors of Ω: |1⟩, |2⟩, |3⟩, and suppose that |Ψ⟩ is a vector in 3 – d
�⃗. Say
�⃗ 𝒋,
�⃗ 𝒌
space. |1⟩, |2⟩, |3⟩ are the unit vectors 𝒊,
If we normalize it, we get
|Ψ⟩ = √2|1⟩ + √3|2⟩ + |3⟩
|Ψ⟩ =
√3
√2
|1⟩ + 2 |2⟩ +
3
1 1 1
√6|3⟩
The probabilities of measuring |1⟩, |2⟩, |3⟩ are , , respectively. No other value is possible.
3 2 6
According to the third postulate, if we make a measurement and measure, say, the second eigenvector,
|2⟩, the state |Ψ⟩ changes and becomes the eigenvector |2⟩.
In terms of the projection operator, the measurement acts like one; here it acted like, ℙ2 , i.e.
the projection operator along the basis vector |2⟩
ℙ2 |Ψ⟩ = |2⟩⟨2|Ψ⟩
If |Ψ⟩ is in the eigenstate |2⟩,
=
√2
|2⟩
2
|Ψ⟩ = |2⟩
Subsequent measurements of the same operator, Ω, will yield the same state |2⟩; repeated applications
of the projection operator don’t have an added effect.
Shoshan 32
Thus, we have three physical states. Each state has a vector that corresponds to it. A
measurement filters out one of those vectors, and that is the result of the measurement.
3.2. The continuous Eigenvalue Spectrum
In the previous example, the eigenvalues were discrete. If they are continuous the probability
needs a different interpretation.
⟨𝜔|Ψ⟩2 can’t be the probability of measuring |𝜔⟩, because there is an infinite number of values
for ω, so each has a probability of zero. Rather, we interpret it to be the probability density. That means
𝑃(𝜔)𝑑𝜔 = ⟨𝜔|Ψ⟩2 is the probability of getting a value between ω and ω + dω.
The two operators 𝑋 and 𝑃 both have continuous eigenvalue spectrums. Earlier, we noted that
while in classical mechanics we only needed two variables 𝑥 and 𝑝, in quantum mechanics we need an
infinite number of variables all contained in the wave function. It is now clear, in classical mechanics a
particle is in a state where it has exactly one definite value. In quantum mechaincs, it is in the state
where it can yield any value upon measurement. Then, we must have probabilites for all those results.
3.3 Collapse of the Wave Function and the Dirac Delta Function
According to postulate III, after a measurement of Ω, the state |Ψ⟩ will collapse to the measured
value |𝜔⟩. Before the measurement, |Ψ⟩ was a linear combination of the eigenvectors of Ω, |𝜔⟩
|Ψ⟩ = �|𝜔𝑖 ⟩ ⟨𝜔𝑖 |Ψ⟩
𝑖
or
|Ψ⟩ = �|𝜔⟩ ⟨𝜔|Ψ⟩𝑑𝜔
After the measurement, it collapses and becomes the measured vector |𝜔⟩
|Ψ⟩ = |𝜔⟩
Suppose we measured the position of the particle. The corresponding operator is 𝑋. Prior to the
measurement we had
Now, we have
|Ψ⟩ = �|𝑥⟩ ⟨𝑥|Ψ⟩𝑑𝑥
|Ψ⟩ = |𝑥′⟩
To find how Ψ(𝑥) looks like, we dot both sides with a basis bra
⟨𝑥|Ψ⟩ = ⟨𝑥|𝑥 ′ ⟩ = 𝛿(𝑥 − 𝑥 ′ )
The collapsed wave function becomes the delta function. This makes sense. When we make a
measurement and find the particle, we expect it to be at the spot we found it, so all the probability is
spiked at that spot.
3.4. Measuring X and P
In the previous example, information was extracted from |Ψ⟩ where the eigenbasis was
discrete. In the following examples, we will deal with continuous eigenbasis.
We have a particle that is in the state |Ψ(𝑥)⟩; the wave function |Ψ(𝑥)⟩ is the Gaussian.
Shoshan 33
Ψ(𝑥) = 𝐴𝑒 −(𝑥−𝑎)
Let us first normalize it.
∞
�
−∞
|Ψ(𝑥)|2
∞
𝑑𝑥 = � 𝐴𝑒 −(𝑥−𝑎)
−∞
∞
𝐼 = � 𝐴2 𝑒 −(𝑥−𝑎)
−∞
We set 𝐼 equal to this integral. Then, 𝐼 2 will be
∞
𝐼 2 = 𝐴4 � 𝑒
−∞
−(𝑥−𝑎)2 ⁄Δ2
We make the substitutions 𝑢 = (𝑥 − 𝑎)⁄Δ,
𝐼
2
4 2
∞
= 𝐴 Δ �𝑒
−∞
−𝑢2
∞
𝑑𝑢 � 𝑒
−∞
2 ⁄2Δ2
2 ⁄Δ2
𝐴𝑒 −(𝑥−𝑎)
2 ⁄2Δ2
𝑑𝑥
𝑑𝑥 = 1
∞
𝑑𝑥 � 𝑒 −(𝑦−𝑎)
−∞
2 ⁄Δ2
𝑑𝑦
𝑣 = (𝑦 − 𝑎)⁄Δ
and
−𝑣 2
2 ⁄2Δ2
4 2
∞
∞
𝑑𝑣 = 𝐴 Δ � � 𝑒 −(𝑢
and 𝑟 2 = 𝑢2 + 𝑣 2
Then, we switch to polar coordinates: 𝑑𝑢 𝑑𝑣 = 𝑟 𝑑𝑟 𝑑𝜃
∞
𝐼 2 = 𝐴4 Δ2 � 𝑟 𝑒
0
−∞ −∞
2 +𝑣 2 )
𝑑𝑣 𝑑𝑢
2𝜋
−𝑟 2
After another substitution, 𝑧 = 𝑟 2 , the integral becomes
𝑑𝑟 � 𝑑𝜃
0
∞
𝐼 2 = 𝐴4 Δ2 𝜋 � 𝑒 −𝑧 𝑑𝑧 = 𝐴4 Δ2 𝜋
0
and
𝐼 = 𝐴2 Δ √𝜋 = 1
1
Thus, 𝐴 = (𝜋Δ2 )−4 and the normalized state is
|Ψ⟩ =
1
1
(𝜋Δ2 )4
𝑒 −(𝑥−𝑎)
2 ⁄2Δ2
Lets say we want to know where the particle is, i.e. what is its position. The operator corresponding to
position is 𝑋. We follow the algorithm we developed. First, we expand |Ψ⟩ in the eigenbasis of 𝑋.
∞
∞
−∞
−∞
|Ψ⟩ = � |𝑥⟩ ⟨𝑥|Ψ⟩ 𝑑𝑥 = � |𝑥⟩ Ψ(𝑥) 𝑑𝑥
Shoshan 34
The probability of finding the particle between 𝑥 and 𝑥 + 𝑑𝑥 is
1
|Ψ(𝑥)|2 𝑑𝑥 =
𝑒 −(𝑥−𝑎)
1
2
(𝜋Δ2 )
2 ⁄Δ2
𝑑𝑥
The particle will probably be found around 𝑎, that is where the probability density is largest.
Next, we look at the momentum of the particle. The operator corresponding to momentum is 𝑃.
In this case the process is longer. We found the eigenvectors last section.
⟨𝑝|Ψ⟩2
1
|𝑝⟩ = Ψ𝑝 (𝑥) =
√2𝜋ℏ
will give us the probability density.
∞
𝑒
𝑖𝑝𝑥
ℏ
∞
⟨𝑝|Ψ⟩ = � ⟨𝑝|𝑥⟩ ⟨𝑥|Ψ⟩ 𝑑𝑥 = � Ψ𝑝∗ (𝑥) Ψ(𝑥) 𝑑𝑥
−∞
−∞
∞
2
2
𝑒 −𝑖𝑝𝑥⁄ℏ 𝑒 −(𝑥−𝑎) ⁄2Δ
𝑑𝑥
(𝜋Δ2 )1⁄4
√2𝜋ℏ
⟨𝑝|Ψ⟩ = �
−∞
To integrate, we expand the powers, add the fractions, and make the following substitutions:
𝑘=
𝛽=
1
√2𝜋ℏ
(𝜋Δ2 )−1⁄4
𝑎ℏ − 𝑖𝑝Δ2
ℏΔ2
𝛼=
𝛾= −
∞
⟨𝑝|Ψ⟩ = 𝑘𝑒 𝛾 � 𝑒 −𝛼𝑥
−∞
Completing the square, we get
𝛽 2 ⁄4𝛼
𝛾
⟨𝑝|Ψ⟩ = 𝑘 𝑒 𝑒
1
2Δ2
∞
�𝑒
−∞
2 +𝛽𝑥
𝑎2
2Δ2
𝑑𝑥
−𝛼�𝑥−
𝛽 2
�
2𝛼
𝑑𝑥
Using the same method we used when we normalized the state we get,
Substituting back the original values,
⟨𝑝|Ψ⟩ = 𝑘 𝑒 𝛾 𝑒 �𝛽
1⁄4
Δ2
⟨𝑝|Ψ⟩ = � 2 �
𝜋ℏ
2 ⁄4𝛼 �
𝜋
𝛼
�
𝑒 (−𝑖𝑝𝑎⁄ℏ) 𝑒 �−𝑝
2 Δ2 ⁄2ℏ2 �
In both of these examples, position and momentum, the eigenvectors reprsented definite
states. When we measure position, |Ψ⟩ changes from being a sum over the states |𝑥⟩ to taking the value
of one of the|𝑥⟩. Before the measurement, the particle had no definite position. It was in a state where
it could yield any |𝑥⟩, i.e. any position, with a certain probability. After the measurement, it is forced
Shoshan 35
into a definite state of position, |𝑥⟩. It is the same with momentum; |𝑝⟩ represents a state of definite
momentum.
For energy, we first have to get the states of definite energy by solving this equation
−
ℏ2 𝑑2 Ψ𝐸
+ 𝑉(𝑥)Ψ𝐸 (𝑥) = 𝐸Ψ𝐸 (𝑥)
2𝑚 𝑑𝑥 2
This is called the time – independent Schrodinger equation. Ψ𝐸 represent states of definite energy. For
position and momentum the definite states are always the same. For energy they depend on the
potential the particle is experiencing.
3.5. Schrodinger’s Equation
We now move on to the fourth postulate. While the first three postulates dealt with the state
|Ψ(𝑥)⟩ at one instant, the fourth deals with how the state |Ψ(𝑥, 𝑡)⟩ evolves with time.
Postulate IV. The wave function |Ψ(𝑥, 𝑡)⟩ follows the Schrodinger equation
𝑖ℏ
𝑑
|Ψ(𝑡)⟩ = 𝐻|Ψ(𝑡)⟩
𝑑𝑡
Schrodinger’s equation is the quantum mechanical counterpart of Newton’s equation 𝐹⃗ = 𝑚
, they both tell you how a particle evolves with time. 𝐻 is the Hamiltonian; it gives the total energy of
the particle
𝐻 =𝑇+𝑉
Where 𝑇 is the kinetic energy and 𝑉 is the potential energy. For every particle the potential, may be
different, so the Hamiltonian may be different.
The Schrodinger equation tells us how the state will change. |Ψ⟩ starts looking perhaps like A
and later like B. If we will solve the Schrodinger equation we will know how it will look.
We develop a solution using two different but equivalent approaches. We first rewrite 𝐻
𝐻 =𝑇+𝑉 =
𝑃2
ℏ2 𝜕 2
+𝑉 = −
+𝑉
2𝑚
2𝑚 𝜕𝑥 2
So, the Schrodinger equation can be written as
𝑖ℏ
𝜕Ψ
ℏ2 𝑑2 Ψ(𝑥)
= −
+ 𝑉Ψ(𝑥)
𝜕𝑡
2𝑚 𝑑𝑥 2
𝑑2 𝑥
𝑑𝑡 2
Shoshan 36
Assuming 𝑉 doesn’t depend on time, we look for solutions of the form
Ψ(𝑥, 𝑡) = Ψ(𝑥)𝑓(𝑡)
𝑑𝑓
𝜕Ψ
= Ψ
𝑑𝑡
𝜕𝑡
𝜕2Ψ
𝑑2 Ψ
=
𝑓
𝜕𝑥 2
𝑑𝑥 2
and
Putting these into Schrodinger’s equation
𝑑𝑓
ℏ2 𝑑2 Ψ
𝑖ℏΨ(𝑥, 𝑡)
= −
𝑓(𝑡) + 𝑉Ψ(𝑥, 𝑡)𝑓(𝑡)
2𝑚 𝑑𝑥 2
𝑑𝑡
Dividing both sides by Ψ𝑓 we get
ℏ2 𝑑2 Ψ 1
1 𝑑𝑓
= −
+𝑉
𝑖ℏ
2𝑚 𝑑𝑥 2 Ψ
𝑓 𝑑𝑡
The left side only depends on 𝑡, and the right side only on 𝑥. This equation is true only if both
sides equal a constant. Otherwise, if we vary either 𝑥 or 𝑡 keeping the other side constant, the equation
will not be true. Thus,
𝑖ℏ
and
−
(3.5.1) Is easily solved
1 𝑑𝑓
=𝐸
𝑓 𝑑𝑡
ℏ2 𝑑2 Ψ 1
+𝑉 =𝐸
2𝑚 𝑑𝑥 2 Ψ
or
or
−
𝑑𝑓
𝐸
= −𝑖 𝑓
𝑑𝑡
ℏ
(3.5.1)
ℏ2 𝑑 2 Ψ
+ 𝑉(𝑥)Ψ(𝑥) = 𝐸Ψ(𝑥)
2𝑚 𝑑𝑥 2
(3.5.2)
𝑓𝑡 = 𝑒 −𝑖𝐸𝑡⁄ℏ
We recognize (3.5.2) as the time independent Schrodinger equation. This tells us that Ψ(𝑥) is in a state
of definite energy. We conclude that the Schrodinger equation
does admit solutions of the form Ψ(𝑥, 𝑡) = Ψ(𝑥)𝑓(𝑡), provided that 𝑓 is the exponential written above
and that Ψ(𝑥) is in a state of definite energy. If these are true then Ψ(𝑥, 𝑡) evolves according to
Ψ(𝑥, 𝑡) = Ψ(𝑥)𝑒 −𝑖𝐸𝑡⁄ℏ
If not, Ψ(𝑥) will be a sum over definite states of energy and evolve in some complex manner.
Ψ(𝑥, 𝑡) = � 𝐴𝐸 Ψ𝐸 (𝑥)𝑒 −𝑖𝐸𝑡⁄ℏ
𝐸
Where 𝐴𝐸 are constants. We now follow a different approach starting with Schrodinger’s equation
Shoshan 37
𝑖ℏ
𝑑
|Ψ⟩ = 𝐻|Ψ⟩
𝑑𝑡
We solve the problem using the propogator. We look for 𝑈(𝑡) that satisfies
We rewrite the Schrodinger equation as
|Ψ(𝑡)⟩ = 𝑈(𝑡)|Ψ(0)⟩
�𝑖ℏ
(3.5.3)
𝑑
− 𝐻� |Ψ⟩ = 0
𝑑𝑡
(3.5.4)
As in the string problem (section 2.16) we need to write the propogator in terms of the eigebasis of 𝐻.
𝐻|𝐸⟩ = 𝐸|𝐸⟩
This is the time independent Schrodinger equation. We expand |Ψ⟩ in the eigenbasis |E⟩
|Ψ(𝑡)⟩ = �|𝐸⟩⟨𝐸|Ψ(𝑡)⟩ = �𝐴𝐸 (𝑡)|𝐸⟩
Where we made the substitution 𝐴𝐸 (𝑡) = ⟨𝐸|𝛹(𝑡)⟩. We operate on both sides with �𝑖ℏ
�𝑖ℏ
𝜕
𝜕𝑡
(3.5.5)
− 𝐻�
𝜕
𝜕
− 𝐻� |𝛹(𝑡)⟩ = � 𝑖ℏ 𝐴𝐸 (𝑡)|𝐸⟩ − 𝐻𝐴𝐸 (𝑡)|𝐸⟩ = ��𝑖ℏ𝐴̇𝐸 (𝑡) − 𝐸𝐴𝐸 � |𝐸⟩
𝜕𝑡
𝜕𝑡
By (3.5.4) the left side is zero, so we have
𝑖ℏ𝐴̇𝐸 = 𝐸 𝐴𝐸
This is easily solved as
𝐴𝐸 (𝑡) = 𝐴𝐸 (0)𝑒 −𝑖𝐸𝑡⁄ℏ
Substituting back 𝐴𝐸 (𝑡) = ⟨𝐸|𝛹(𝑡)⟩, and plugging it into (3.5.5)
⟨𝐸|𝛹(𝑡)⟩ = ⟨𝐸|𝛹(0)⟩𝑒 −𝑖𝐸𝑡⁄ℏ
|𝛹(𝑡)⟩ = �|𝐸⟩⟨𝐸|𝛹(0)⟩ 𝑒 −𝑖𝐸𝑡⁄ℏ
By comparing this to (3.5.3) we see that
𝑈(𝑡) = �|𝐸⟩⟨𝐸| 𝑒 −𝑖𝐸𝑡⁄ℏ
𝐸
Shoshan 38
4. Particle in a Box
We now consider a situation that features one of the amazing results of Quantum Physics. We
have a particle between 0 and 𝑎 in a potential 𝑉(𝑥) so that
𝑉(𝑥)
𝑉(𝑥) = �
0,
∞,
0≤𝑥<𝑎
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
The particle can’t escape because it needs an infinite amount of energy. We look
for states of definite energy, so we use the time independent Schrodinger equation.
−
ℏ2 𝑑2 Ψ
= 𝐸Ψ(𝑥)
2𝑚 𝑑𝑥 2
𝑑2 Ψ
= −𝑘 2 Ψ
𝑑𝑥 2
𝑘≡
The solution to this equation is
0
√2𝑚𝐸
ℏ
Ψ(𝑥) = 𝐴 sin 𝑘𝑥 + 𝐵 cos 𝑘𝑥
Where the potential is infinte Ψ(𝑥) = 0. Then, the wave function will be continuous only if
Ψ(0) = Ψ(𝑎) = 0
Ψ(0) = 𝐴 sin(0) + 𝐵 cos(0) = 𝐵 = 0
Ψ(𝑎) = 𝐴 sin 𝑘𝑎 = 0
If 𝐴 is to be nonzero, we must have
𝑘𝑎 = 𝜋, 2𝜋, 3𝜋, ⋯
Or
𝑘=
𝑛𝜋
𝑎
We substitute back for 𝑘 from (4.1) and solve for 𝐸
Replacing 𝑘 with the values we got,
𝑛 = 1, 2 ,3, ⋯
𝐸=
𝐸𝑛 =
𝑛2 𝜋 2 ℏ2
2𝑚𝑎2
ℏ2 𝑘 2
2𝑚
𝑛 = 1, 2, 3, ⋯
Thus, the definite states of energy can’t have just any energy. Energy is quantized to certain values.
a
(4.1)
Shoshan 39
Conclusion
In the days of classical physics everything was predictable. Given all the forces and
momenta of all the particles in a system, the entire future of the system can theoretically be
determined. In quantum physics however, nothing can be determined definitely.
Consider a simple idealized example. We have a particle in a closed box ( a real box, not
the one from Section 4) that can have only one of two positions A or B. In classical physics we
could say the particle is either definitely in A or definitely in B. If we then open the box and see
it in A, we know it was in A even prior to the measurement.
It is not so in quantum physics; when the box is closed the particle is not in A or in B. It is
in the state that upon measurement it can be found either in A or B. There is a common
misconception that when the box is closed the particle is partly in A partly in B. It is not. It is in
this bizarre quantum state where it doesn’t make sense to ask “where is the particle?”
When we open the box and find it in A, it does not mean it was always in A. Rather, it
means that upon measurement, the particle was forced into position state A. Meaning, before
the measurement it had a probability of being found in A or in B, say 50/50. When the
measurement occurs the particle suddenly finds itself in say A. Before the measurement, the
particle itself doesn’t know where it will end up. So we can’t make a definite prediction as in
pre-quantum times. We can only say it has a 50% chance of being in A.
This may sound very weird, but that is only because we experience a Newtonain world,
not a quantum world. Mass plays a role here, the more mass a particle has the less the wave
function of the particle has an effect. The objects we deal with everyday are so enormously
massive, compared to electrons, that we see no quantum effects.
If there were quantum people, i.e. people with the mass of electrons, they may be just
as amazed with our Newtonian world. They will be astonished at how we can make such
accurate predictions using Newton’s laws.
Richard Feynman once said, “No one understands quantum mechanics.” Perhaps, that is
what makes the subject so fascinating.
Shoshan 40
Bibliography
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University Press, 2006
Borowitz, Sidney. Fundamentals of Quantum Mechanics: Particles, Waves, and Wave
Mechanics.Benjamin, 1967.
Feynman, Richard, R. Leighton and M. Sands. The Feynman Lectures on Physics Volume 3. Addison –
Wesley, 1964
Griffiths, David J. Introduction to Quantum Mechanics. New Jersey: Prentice Hall, 1995.
Shankar, Ramamurti. Principles of Quantum Mechanics. New York: Plenum Press, 1980.