Physics 115A Review
Luca Niu
June 14, 2023
Contents
1 Important Formula
1.1 Gaussian Integral . . . . . . . . .
1.2 Commutator . . . . . . . . . . .
1.3 Adjoint Operator . . . . . . . . .
1.4 General Property of Bound State
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3
3
3
3
3
2 Harmonic Oscillator (Algebraic)
2.1 The Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Raising and Lowering Operator . . . . . . . . . . . . . . . . . . . . . . .
2.3 Hamiltonian Operator in Terms of â± . . . . . . . . . . . . . . . . . . .
2.4 Constructing Full Spectrum of Solution . . . . . . . . . . . . . . . . . .
2.4.1 Lower and Raising of Energy . . . . . . . . . . . . . . . . . . . .
2.4.2 The Ground State . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3 All Allowed Energy . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.4 Normalized Higher Energy State . . . . . . . . . . . . . . . . . .
2.5 Problem Solving Tricks with Raising and Lowering Operators . . . . . .
2.5.1 Prove the Orthogonality of Stationary States . . . . . . . . . . .
2.5.2 Find Expectation Values of Some Quantity of a Stationary State
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4
4
4
4
5
5
5
5
5
6
6
6
3 Free Particle
3.1 Potential and Schrodinger Equation . . . . .
3.2 Stationary Solution and Boundary Condition
3.3 General Solution . . . . . . . . . . . . . . . .
3.4 Procedure for Problem Solving . . . . . . . .
3.5 Group Velocity and Phase Velocity . . . . . .
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7
7
7
7
8
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4 Delta Potential
4.1 Bound State and Scattering State . . . . . . . . . . . . .
4.2 The Potential . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Bound State . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Stationary State Solution . . . . . . . . . . . . .
4.3.2 Restriction by Boundary Condition . . . . . . . .
4.4 Scattering State . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 General Solution . . . . . . . . . . . . . . . . . .
4.4.2 Boundary Condition (Constraint on Coefficients)
4.4.3 Transmission and Reflection . . . . . . . . . . . .
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8
8
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9
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10
10
10
10
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Cases
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11
11
11
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12
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13
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5 Finite Potential Well
5.1 Potential . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Bound State . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Solve in Each Region . . . . . . . . . . . . . .
5.2.2 Even Solution . . . . . . . . . . . . . . . . . .
5.2.3 Interpretation of the Constraint and Limiting
5.3 Scattering State . . . . . . . . . . . . . . . . . . . . .
1
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6 Hilbert Space
6.1 Definition of Hilbert Space . . . .
6.2 Inner Product Notation . . . . .
6.3 Orthogonality . . . . . . . . . . .
6.4 Normal . . . . . . . . . . . . . .
6.5 Complete . . . . . . . . . . . . .
6.6 Analogous between Hilbert Space
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14
14
14
14
14
14
14
. . . . . . . . . . . . . .
to Hermitian Operator .
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14
14
15
15
15
8 Eigenfunction of Operators
8.1 Discrete and Continuous Spectrum . . . . . . . . . . . . . . . . . . . . . . .
8.2 Discrete Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Eigenvalues are Real . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.2 Eigenfunctions Associated with Different Eigenvalues are Orthogonal
8.2.3 Eigenfunctions are Complete . . . . . . . . . . . . . . . . . . . . . .
8.3 Continuous Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.1 Eigenfunction of Momentum Operator . . . . . . . . . . . . . . . . .
8.3.2 Eigenfunction of the Position Operator . . . . . . . . . . . . . . . . .
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15
15
15
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16
16
16
16
17
9 Generalized Statistical Interpretation
9.1 Probability of Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Position and Momentum Space Wavefunction from Generalized Perspective . . . . . . . . . . . .
17
17
17
18
10 Uncertainty Principle
10.1 Generalized Uncertainty Principle . .
10.2 Energy Time Uncertainty Principle .
10.2.1 Statement and Interpretation
10.2.2 Proof . . . . . . . . . . . . .
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18
18
19
19
19
11 Change of Basis
11.1 Generalized Wave Function
11.2 Wavefunction as Vectors . .
11.3 Operator as Matrix . . . . .
11.4 Projection Operator . . . .
11.5 Identity Operator . . . . . .
11.6 Change of Basis . . . . . . .
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20
20
21
21
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21
22
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and Vector Space .
7 Observation
7.1 Hermitian Operator . . . . . . . .
7.2 Observable Quantities Corresponds
7.3 Hermitian Conjugate . . . . . . . .
7.4 Determinate State . . . . . . . . .
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1
Important Formula
1.1
Gaussian Integral
Even Functions:
∞
Z
e
−ax2
0
∞
Z
1
=
2
2
x2 e−ax =
0
Odd Function:
Z
Z
∞
r
1
4a
2
x3 e−ax =
0
π
a
1
2a
2
x3 e−ax =
0
∞
π
a
r
1
2a2
Note that we are here integrating on the half line, for wave function we usually integrate around the whole line.
We are going to use the even or odd property to determine their integral on the whole line
1.2
Commutator
[Â, B̂] = ÂB̂ − B̂ Â
1.3
(1)
Adjoint Operator
Suppose â− and â+ are adjoint operators, then for any function f and g (vanish at infinity)
Z ∞
Z ∞
f ∗ (â± g)dx =
(â∓ f )∗ gdx
−∞
(2)
−∞
We are using the standard notion of the inner product of functions on whole line here
The Hermitian conjugate (adjoint) says that applying the operator to the second function is equivalent to
applying the conjugate of that operator to the first function
1.4
General Property of Bound State
• For bound state (normalizable stationary state solution), the separation constant E must be real
Proof: Write it in complex form, and show the imaginary part has to be 0 by normalization
• Solutions to time independent Schrodinger equation can always be expressed in real form (i.e.If we get a
complex solution, we can always construct a combination of real solutions with the same energy, take this
as sort of an equivalence, but does not mean we can reproduce the complex solution literally)
Proof: Suppose ψ is a complex solution, then its complex conjugate ψ ∗ must also solves the (time
independent) solution. By linearity, the combination ψ + ψ ∗ and i(ψ − ψ ∗ ) are also solutions, and they
are real
• If V (x) is even, then the stationary state solution is either even or odd (connect this to when solving the
finite potential well, we separated the even and odd solutions)
Proof: If V (x) is even and ψ(x) is a solution to this potential, then V ψ(−x) is also a solution. Then
ψ(x) ± ψ(−x) are even and odd solution with the same energy respectively
• E cannot be lower than the minimum potential of the entire setup
• If there are two distinct solutions with the same energy, we said they are degenerate solutions. (e.g. For
free particle we have a wave traveling to left and write both solution to the same energy). For bound
state, there is no degenerate solutions
Proof: Suppose ψ1 and ψ2 are two solutions, multiply the Schrodinger equation for ψ2 by ψ1 , the subtract
two equation and show that the two solutions are in fact the same
• The number of nodes of stationary states of 1D potential always increase with energy
Proof: Suppose ψn has higher energy than ψm , then show between each node of ψm there is (at least) a
node of ψn . Having node in a region means it changes sign for some times in that region
3
2
Harmonic Oscillator (Algebraic)
2.1
The Potential
Harmonic oscillator has potential of the form
V (x) =
1
mω 2 x2
2
The Time independent Schrodinger equation can be written as
−
2.2
ℏ2 ∂ 2 ψ 1
+ mω 2 x2 ψ = Eψ
2m ∂x2
2
(3)
Raising and Lowering Operator
We define following two operators:
(
â+ =
â− =
√ 1
(−ip̂
2ℏmw
√ 1
(+ip̂
2ℏmw
+ mwx)
+ mwx)
(4)
Note that the definition is just different in terms of the sign before ip̂ which is the only term involving i, we do
this because we want them to be Hermitian conjugate of each other
2.3
Hamiltonian Operator in Terms of â±
Now we want to write the Hamiltonian operator in terms of raising and lowering operators
We want to do this because:
• It can be used to show that these operator helps us to construct the solution of the whole energy spectrum
from the solution of some particular energy
• It can help us to formulate an explicit equation that determines the normalization constant of all energy
states
Firstly, let’s find out what is the Hamiltonian operator in this case in terms of position (x) and momentum
∂
operator (−iℏ ∂x
)
Rewrite the left side of the Schrodinger equation
2
1
∂ ψ
2 2 2
ℏ 2 + m ω x ψ = Eψ
2m
∂x
1
(p̂2 + (mω)2 x2 )ψ = Eψ = Ĥψ
2m
Now since the Hamiltonian operator contains square of position and momentum operator, let’s multiply â− and
â+ together which will also give us such squares
â− â+ =
1
(p̂2 + (mω)2 x2 − imω(xp̂ − p̂x))
2ℏmω
1
i
(p̂2 + (mω)2 x2 ) −
[x, p̂]
2ℏmω
2ℏ
By applying the required operator combination to a test function, we get
=
[x, p̂] = iℏ
Thus,
1
1
Ĥ = ℏω(â− â+ − ) = ℏω(â+ â− + )
2
2
The last identity is derived from the following equation
[â− , â+ ] = 1
4
(5)
(6)
2.4
2.4.1
Constructing Full Spectrum of Solution
Lower and Raising of Energy
Theorem:
If wave function ψ is the solution to the harmonic oscillator at energy E, then â+ ψ is the solution to energy
E + ℏω, similarly, â− ψ is the solution to energy E − ℏω
Proof :
We know the compact form of time independent Schrodinger equation Ĥψ = Eψ, thus if we apply the
Hamiltonian operator to the wave function, its energy should appear as a coefficient of that wave function,
thus
• If we apply the Hamiltonian operator Ĥ = ℏω(â− â+ − 12 ) to â− ψ, we will get that it is equal to
(E − ℏω)(â− ψ)
• If we apply the Hamiltonian operator Ĥ = ℏω(â+ â− + 12 ) to â+ ψ, we will get that it is equal to
(E + ℏω)(â+ ψ)
The trick involved in this proof is to use the commutation relation [â− , â+ ] = 1, we can switch the order of the
lowering and raising operator in the product, then we get an additional 1 term, which allows us to use the fact
that ψ solves the equation with energy E, i.e. Ĥψ = Eψ
Also, remember that the operators always commute with constants
2.4.2
The Ground State
For the ground state, if we apply lowering operator to it, we should get 0, which is a trivial solution to the time
independent Schrodinger equation, but a non-normalizable one, so the ground state is the lowest physical state.
i.e.
d
1
√
ℏ
+ mωx ψ0 = 0
dx
2ℏmω
Rearrange the equation and do an integral on both sides, we get
mω
2
ψ0 = Ae− 2ℏ x , A = (
mω 1
)4
πℏ
(7)
A is the normalization constant determined separately
2.4.3
All Allowed Energy
We just need to determine the ground state energy, then since by applying the raising operator we raise energy
by ℏω, we get all the allowed energy
To determine the ground state energy, apply the Hamiltonian operator expressed in the form Ĥ = ℏω(â+ â− + 21 )
to the ground state eave function ψ0 Since â− ψ0 = 0 (which is why we want Ĥ to be in this form), we get
E0 =
2.4.4
1
1
ℏω =⇒ En = ( + n)ℏω
2
2
(8)
Normalized Higher Energy State
Given the wave function ψn of some particular energy, we can find solution of both higher and lower energy by
applying â± , but for wave function of these energy, we need normalized solution, so we want a convenient and
explicit expression for the normalization constants
Suppose
(
cn ψn+1 = â+ ψn
dn ψn−1 = â− ψn
where ψn+1 and ψn−1 are also wave functions
To find the normalization constant, we integrate the solution of the higher and lower energy state, we also apply
the property of Hermitian conjugate for later convenience
Z ∞
Z ∞
Z ∞
2
2
∥cn ∥
∥ψn+1 ∥ dx =
(â+ ψn )∗ (â+ ψn )dx =
(â− â+ ψn )∗ (ψn )dx
−∞
−∞
−∞
5
2
Z
∞
Z
2
∞
(â− ψn )∗ (â− ψn )dx =
∥ψn−1 ∥ dx =
∥dn ∥
−∞
−∞
Z
∞
(â+ â− ψn )∗ (ψn )dx
−∞
Now the multiplication of these operators acting on ψn is actually very easy to calculate
By equation (5), and written energy En of the state explicitly, we have
(
Ĥψn = ℏω(â+ â− + 21 )ψn = Eψn = ℏω( 12 + n)ψn
Ĥψn = ℏω(â− â+ − 12 )ψn = Eψn = ℏω( 12 + n)ψn
Thus,
(
â+ â− ψn = nψn
â− â+ ψn = (n + 1)ψn
(9)
Plug this back into the integral to find the normalization constants
Z ∞
Z ∞
2
2
2
∥ψn+1 ∥ dx = (n + 1)
∥ψn ∥ dx
∥cn ∥
−∞
2
Z
−∞
∞
Z
2
∞
∥ψn−1 ∥ dx = n
∥dn ∥
−∞
2
∥ψn ∥ dx
−∞
Using the fact that both all wave functions are normalized,
(
√
cn = n + 1
√
dn = n
Thus, our final solution for the n-th energy state is
1
ψn = √ (â+ )n ψ0
n!
(10)
ψ0 is already normalized so we just need to find the effect of applying the raising operator.
The n + 1 in the expression of cn may be confusing, but note that this n + 1 aligns with ψn+1 in the proof which
is the
√ index of the wavefunction we get by applying the raising operator. Rewrite the index as n, we would get
an n! in the denominator
2.5
2.5.1
Problem Solving Tricks with Raising and Lowering Operators
Prove the Orthogonality of Stationary States
We can prove the orthogonality of stationary states of harmonic oscillator by applying an operator â+ â− to
the inner product, because
Z ∞
Z ∞
∗
∗
ψm (â+ â− )ψn dx = n
ψm
ψn dx
−∞
−∞
by applying these operator in the original order, however, we can also use the Hermitian conjugate property
Z ∞
Z ∞
Z ∞
∗
∗
ψm
(â+ â− )ψn dx =
(â+ â− ψm )∗ ψn dx = m
ψm
ψn dx
−∞
−∞
−∞
Thus,
Z
∞
∗
ψm
ψn dx = 0
(m − n)
−∞
If m ̸= n, then the inner product has to go to 0 (note that the definition for inner product of functions includes
the integral already)
2.5.2
Find Expectation Values of Some Quantity of a Stationary State
We know the operators of all dynamic quantities can be written as a function of position and momentum
operators, we write the position and momentum operator as a a function of â±

q
ℏ
x =
(â+ + â− )
q2mω
(11)
p̂ = i ℏmω (â − â )
+
2
−
Then we can apply the operator Q̂ of some quantity written in terms of â± applied to the inner product. We
can exploit the orthogonality condition and relation in equation (9) to calculate the expectation value of that
quantity
6
3
3.1
Free Particle
Potential and Schrodinger Equation
For a free particle, the potential is V (x) = 0, i.e. 0 everywhere
Thus, the Schrodinger equation reads:
ℏ2 ∂ 2 ψ
−
= Eψ
(12)
2m ∂x2
Combine all the constants, we get
√
ℏ2 ∂ 2 ψ
2mE
2
= −k ψ, k =
(13)
2
2m ∂x
ℏ
Why do we know that E is certainly positive? We know that the energy cannot be lower than the potential
in the entire setup, otherwise there is no normalizable solution. In this case the potential is always 0, so the
energy has to be greater than 0 (Energy being 0 violates the uncertainty principle)
3.2
Stationary Solution and Boundary Condition
The equation is the same as inside the infinite square well, and it seems that we should have the same solution.
The solution for a particular state is indeed the same, however, since the free particle does not have any boundary
condition that restricts the value of k (i.e. allowed energy). The energy can take any value, and there is no
normalizatble stationary state with definite energy.
Thus, we write the solution as complex exponential, because sinusoidal form allows us to impose boundary
condition better, but complex exponential help us to do Fourier Transform more conveniently
ψ(x) = Aeikx + Be−ikx
(14)
−iE
ℏ t
(simply multiply it to the time independent solution and reorganize the
Introduce the time part solution e
equation), we get full solution for a (pseudo) stationary state
ℏk
ℏk
ψ(x, t) = Aeik(x− 2m t) + Be−ik(x+ 2m t)
(15)
The first term represents a wave of unchanging shape traveling to the right, while the second term travels to
the left. In both cases, k can only take positive values, now we can let k range between all real numbers
(√
k=
2mE
ℏ
√
− 2mE
ℏ
And the (pseudo) stationary state solution written as
ℏk2
ψ(x, t) = Aei(kx− 2m t)
3.3
General Solution
Since k can take a continuous spectrum of value, the coefficients are also captured by a continuous function
ϕ(k), we do a Fourier transform (including time term) of ϕ(k) to find the general solution
Z ∞
ℏk2
1
Ψ(x, t) = √
ϕ(k)ei(kx− 2m t) dk
(16)
2π −∞
We can find ϕ(k) by the inverse Fourier transformation of the initial function ψ(x, 0)
Z ∞
1
ϕ(k) = √
Ψ(x, 0)e−ikx dx
2π −∞
We can do this because we still have orthogonality condition for complex exponential
Z ∞
′
eikx e−ik x dx = 2πδ(k − k ′ )
(17)
(18)
−∞
′
Thus, by doing an inner product with e−ik x , we are still selecting out the coefficients ϕ(k ′ ), just like the discrete
Fourier series case
7
3.4
Procedure for Problem Solving
• Given the initial function, we first normalize it
• Do an inverse transform to find ϕ(k)
• Plug into the integral and integrate to find the general solution
3.5
Group Velocity and Phase Velocity
Suppose we have a wave packet
1
f (x) = √
2π
Z
∞
ϕ(k)ei(kx−ωt) dk
−∞
We assume that the ϕ(k) is peaked at some particular k0 , otherwise, it the momentum composition is too
diverse, the wave packet will dissemble very quickly, there is no well defined notion of group velocity.
Then we can do a Taylor expansion of ω(k) around k0
ω(k) ≈ ω0 + ω0′ (k − k0 )
The lower index of ω denote its evaluation at k0 Plug back into the wave function (of course it does not have
to be in quantum mechanics), let s = k − k0 and rearrange the equation
Z ∞
′
1
ϕ(k0 + s)eis(x−ω0 t) ds
(19)
f (x) = √ ei(k0 x−ω0 t)
2π
−∞
The exponential outside integral represents the wave ripple, we can see it is a wave traveling with speed ω0 /k
The exponential inside the integral represents the wave packet, it is a wave traveling at speed ω0′ , which is the
derivative with respect to k
Thus,
(
vphase = ωk
(20)
vgroup = dω
dk
4
Delta Potential
4.1
Bound State and Scattering State
To distinguish between the bound state and scattering state, we compare the energy with potential at infinity
• If E < V (+∞) and E < V (−∞)
We have a bound state, for bound state there is discrete allowed energy
• If E > V (+∞) or E > V (−∞)
We have a scattering state, all energy values are allowed
Some potential permit only bound state (infinite square well, harmonic oscillator), some potential permit only
scattering state (free particle), some permits both (delta potential, finite square well)
4.2
The Potential
V = −αδ(x)
The shape of the potential is just a spike at x = 0
The negative sign here is very important, because as we mentioned previously, particle cannot have energy
lower than the potential anywhere, thus, to get a bound state, which means the particle has energy less than 0
in this set up, and energy not less than potential anywhere, we must have a negative spike!!
8
4.3
4.3.1
Bound State
Stationary State Solution
For delta potential, the potential is continuous everywhere except x = 0, so we can solve it at other region
except at x = 0, then impose a boundary condition at this point to get full solution
For finite square well which will be discussed later, however, we must find solution inside the well
For bound state, we have E < 0, because that’s the potential at infinity
On the whole real line except x = 0,the Schrodinger equation reads as in the free particle case because V = 0
−
ℏ2 ∂ 2 ψ
= Eψ
2m ∂x2
However, since the energy is negative, in order to combine constants, we have
√
−2mE
k=
ℏ
Thus, we get an exponential decay equation which is different from the free particle case
d2 ψ
= k2 ψ
dx2
(21)
The general solution for a stationary state (for bound state real stationary state with definite energy exists) is
ψ(x) = Aekx + Be−kx
4.3.2
Restriction by Boundary Condition
Firstly, the wave function cannot blow up at infinity Thus, for x < 0 the negative exponential term goes to 0,
for x > 0 the positive exponential term goes to 0, now we get regional solution
(
ψ(x) = F e−kx , x > 0
(22)
ψ(x) = Aekx , x < 0
Now we still haven’t find the constraint on k, which will give us allowed energy and is essential for a
bound state solution
We still have one boundary we haven’t use, namely the x = 0. At this point, ψ(x) should be continuous,
this gives us F = A
dψ
dx
is not continuous because the potential is infinite at x = 0. However we can evaluate the discontinuity
at x = 0 by selecting an arbitrarily small positive real number ϵ
dψ
|x=ϵ = lim+ −kF e−kx = −kF
dx
x→0
dψ
|x=−ϵ = lim kF ekx = kF
dx
x→0−
Thus,
dψ
dψ
|x=ϵ −
|x=−ϵ = −2kF
dx
dx
Delta function has nice integral property, namely integrating a region including it will yields a constant, and
we are integrating the (time independent) Schrodinger equation in an interval [−ϵ, ϵ] which contains 0 to get an
expression for the discontinuity of dψ
dx in a expression without k, this will give us constraint on k
Z
ϵ
(−
−ϵ
ℏ2 d2 ψ
)
dx +
2m dx2
Z
ϵ
Z
ϵ
V ψ(x)dx =
−ϵ
Eψ(x)dx
−ϵ
The right side is 0 because ψ is continuous at x = 0, thus
Z
Z
dψ
dψ
2m ϵ d2 ψ
2m ϵ
2mα
|x=ϵ −
|x=−ϵ = 2
dx
=
−αδ(x)ψ(x)dx = − 2 ψ(0)
dx
dx
ℏ −ϵ dx2
ℏ2 −ϵ
ℏ
9
Since ψ(x) is continuous we can evaluate it at x = 0, which is equal to F , thus
√
mα
−2mE
k= 2 =
ℏ
ℏ
(23)
Thus, the bound state solution is
ψ(x) = F e−k|x| =
r
mα −mα|x|/ℏ2
e
ℏ2
(24)
The constant is determined by normalization. As we can see, there is only 1 allowed energy
4.4
4.4.1
Scattering State
General Solution
For scattering state, we have E > 0, thus although we have exactly the same Schrodinger equation, we redefine
k as
√
2mE
k=
ℏ
such that the rearranged equation is the same as equation (13)
However, we have two solutions corresponding to region x > 0 and x < 0 respectively
(
ψ(x) = F eikx + Ge−ikx , x > 0
(25)
ψ(x) = Aeikx + Be−ikx , x < 0
4.4.2
Boundary Condition (Constraint on Coefficients)
These solution are oscillating and automatically does not blow up at infinity
Firstly, we apply the continuity of wave function at x = 0
A+B =F +G
(26)
Then, since we have evaluated the discontinuity of dψ
dx at x = 0 previously, now we want it to be expressed by k
(
dψ
dx |x=ϵ = ik(F − G)
(27)
dψ
dx |x=−ϵ = ik(A − B)
Since ψ(0) = A + B = F + G, by our previous evaluation
dψ
dx |x=ϵ
ik(F − G − A + B) = −
−
dψ
dx |x=−ϵ
= − 2mα
ℏ2 ψ(0)
2mα
(A + B)
ℏ2
Now we write it in a more useful form by letting
α
mα
β= 2 =
ℏ k
ℏ
r
2m
E
F − G = A(1 + 2iβ) − B(1 − 2iβ)
4.4.3
(28)
Transmission and Reflection
We should recognize that the 4 terms in the solution are traveling waves:
• Aeikx , Be−ikx are wave in the left region traveling to right and left respectively
• F eikx , Ge−ikx are wave in the left region traveling to right and left respectively
(More suggestive form will be multiplying the time part, we will see the traveling direction)
Although not too many constraint are available for the scattering state, in an real experiment, usually a beam
incident from one direction only, suppose it is from left to right. Then there is no incident beam on the right
region, i.e. G = 0
10
Combine equation (28) and (26), we can find
iβ
1
A, F =
A
1 − iβ
1 − iβ
We define the transmission and reflection coefficients as the square of the amplitude of the transmitted wave
and reflected wave dividing that of the incident wave, respectively, so
B=
T =
∥F ∥
∥A∥
2
2
=
1
1 + β2
(29)
=
β2
1 + β2
(30)
2
R=
∥B∥
∥A∥
2
The limiting cases are that when the energy is very high, we should have dominant transmission, while if the
energy is very low, we should have dominant reflection, which is confirmed by our formula because we have
shown as E increases, β decreases, and when β → 0, we get T → 1 and R → 0
Note that both T and R are dependent on α2 , thus if we have a positive spike (i.e. replace α by −α), we
would have no bound states, but the transmission and reflection coefficient stay the same
5
Finite Potential Well
5.1
Potential
(
V =
−V0 , −a < x < a
0, |x| > a
Setting potential to 0 outside the well give us easy solution for the scattering state
5.2
5.2.1
Bound State
Solve in Each Region
For bound state we have E < 0, this makes our k defined the same as the delta bound state, i.e. equation (23)
Similarly to the delta potential problem, we try to solve solution in each region then impose boundary condition
• Outside the well (|x| > a)
We have the same solution for Schrodinger equation as the bound state of delta potential
ψ(x) = Ae−kx + Bekx
With boundary condition that the wave function cannot blow up at infinity imposed
(
Bekx , x < −a
ψ(x) =
F e−kx , x > a
(31)
• Inside the well (−a < x < a)
The time independent Schrodinger equation becomes
−
ℏ2 d2 ψ
− V0 ψ = Eψ
2m dx2
Move the potential term to the right and combine all the constants
∂2ψ
= −l2 ψ
∂x2
where
(32)
p
2m(E + V0 )
(33)
ℏ
Note that l has the same form as the k for the infinite square well and the scattering state of the delta
potential just that E is subtracted the potential −V0 , because in these cases E − V > 0
l=
Then we should have the oscillation solution
ψ(x) = C sin(lx) + D cos(lx)
11
(34)
5.2.2
Even Solution
For even solution, we choose the cosine terms of the solution inside the well.
Since the solution is even, we just need explicit expression on half line, so we can write the solution in the more
suggestive form

−kx

, x>a
F e
ψ(x) = D cos(lx), 0 < x < a
(35)


ψ(−x)
This form is more suggestive because it suggests that we just need half the boundary condition to get all we
can do with the entire set of boundary conditions
Apply boundary conditions that ψ(x) and dψ
dx are continuous at x = a
We will skip the step as there is no new quantity involved and the calculation is straightforward (the only
essential step is to divide the two equations of the boundary condition), we get
k
= tan(la) (= tan(z))
l
(36)
Now our equation involve both k and l, since they are both function of energy, we can use equation (36) to
find constraint on energy.
We have another important correlation between k, l, and E
K 2 + l2 =
2mV0
z2
(= 02 )
2
ℏ
a
(37)
We introduce two convenient notation for combining equation (36) and (37) more easily
z = la,
z0 =
ap
2mV0
ℏ
The choice is for obvious reason because z is the argument in the right side of equation (36) while z02 gives the
right side of equation (37) multiplied by a2
Divide both sides of equation (37) by l2 , then use equation (36), we get the relation
r z0 2
tan(z) =
+1
z
(38)
This equation gives the constraint on energy
5.2.3
Interpretation of the Constraint and Limiting Cases
To get a better grasp on what the constraint on the energy means, we plot the function on both sides of equation
(38), and each intersection represents an allowed energy
We can see that at z = z0 the function on right hand side intersects with the z-axis, this observation allow us
to consider some limiting cases
• When the potentialqwell is extremely deep or wide (z0 very large)
z0 2
We push tan(z) =
+ 1 upward and slide z0 to the right
z
The end point is very far from origin, meaning there are many allowed energy states
12
The intersections are very high, meaning intersections occur very close to zn = (nπ/2) where n is odd
Converting z = la back into the expression involving E + V0 , we will get that E + V0 which is the
energy above potential (which is just E in the case of infinite square well), is exactly the allowed energy
for the infinite square well with odd n (The odd solution at deep or wide potential gives the even energies)
• When the potential well is extremely shallow or narrow (z0 very small)
Push z0 toward the origin, we get that there is just one allowed energy
5.3
Scattering State
The procedure of solving for the scattering state is pretty much the same as that for the delta potential.
• Write down solution in all three region
• Assuming no incident wave from the right (take F = 0)
• Impose boundary condition that both ψ and
dψ
dx
are continuous at x = −a and x = a
• Write the transmission coefficient, which is provided in the book, the reflection coefficient is just 1− the
transmission coefficient
For finite square well, we get
1
V02
2a p
=1+
sin2 ( ) 2m(E + V0 )
T
4E(E + V0 )
ℏ
When sin is made 0 we get perfect transmission, we get that this happens exactly when E + V0 is the allowed
energy for the infinite square well (with width 2a)
13
6
Hilbert Space
6.1
Definition of Hilbert Space
Hilbert space consists of all the square-integrable functions (i.e. the integral of the norm square of the function
is finite).
This condition guarantee that the inner product of any two functions from the Hilbert space is finite by the
Schwartz inequality
s
Z b
Z b
Z b
2
2
∗
f (x)g(x)dx ≤
∥f (x)∥ dx
∥g(x)∥ dx
(39)
a
6.2
a
a
Inner Product Notation
The inner product of two functions f and g can be represented as follows
Z
⟨f |g⟩ =
b
f ∗ gdx
(40)
a
In particular
⟨f |g⟩ = ⟨g|f ⟩
∗
(41)
We can show this by writing the functions into real and complex parts
6.3
Orthogonality
Two functions are orthogonal if their inner product is 0
6.4
Normal
A function is normalized when its inner product with itself is 1
6.5
Complete
A set of functions is complete when any function (in the Hilbert) space can be represented by the linear
combinations of functions in this set
6.6
Analogous between Hilbert Space and Vector Space
• Wavefunction ⇐⇒ Vector
This analogy helps illustrate the projection of wavefunciton into different basis
• Eigenfunction ⇐⇒ Unit Vector
This analogy helps illustrate the expansion of functions and operators in different basis (which can be the
set of eigenfunctions of a particular operator)
• Expansion Coefficient ⇐⇒ Projection on Unit Vector
This analogy helps illustrate how to find coefficient when we expand the wavefunction in a particular basis
7
7.1
Observation
Hermitian Operator
An operator Q̂ is Hermitian if for any function f and g
D
E D
E
f Q̂g = Q̂f g
14
(42)
7.2
Observable Quantities Corresponds to Hermitian Operator
A physical observable must have a real expectation value
The expectation value can be written as
Z
D
E
⟨Q⟩ = Ψ∗ Q̂Ψdx = Ψ Q̂Ψ
Since ⟨Q⟩ is real, it must be equal to its complex conjugate, by equation (41), we know the complex conjugate
is
D
E∗ D
E
∗
⟨Q⟩ = Ψ Q̂Ψ = Q̂Ψ Ψ
Thus, the operator for observables are Hermitian (in a weak sense, because we don’t require two function f and
g)
7.3
Hermitian Conjugate
The Hermitian conjugate of Q̂, denote Q̂† is an operator such that
E
D
E D
f Q̂g = Q̂† f g
In particular for Hermitian operators Q̂ = Q̂†
7.4
Determinate State
Determinate states of an operator Q̂ is a state that every measurement by Q̂ gives the same value
Theorem
Determinate states of Q̂ are eigenstates of Q̂
Proof
Determinate state means that the uncertainty of observation is 0
Suppose that Ψ is a determinate state of Q̂, with definite measurement result q
D
E D
E
σQ = (Q − ⟨Q⟩)2 = Ψ (Q̂ − ⟨Q⟩)2 Ψ = Ψ (Q̂ − q)2 Ψ
Since Q̂ is Hermitian, so must be Q̂ − q, we can move one of it in the square to the front
D
E
σQ = (Q̂ − q)Ψ (Q̂ − q)Ψ = 0
The only function whose inner product with itself is 0 is the 0 function
Thus,
Q̂Ψ = qΨ, q ∈ R
This is the equation for an eigenstate
8
8.1
Eigenfunction of Operators
Discrete and Continuous Spectrum
A spectrum of an operator is the collection of all its eigenvalues
Discrete spectrum are normalizable thus represents physical states, while continuous spectrum are not
Some operator has only discrete spectrum (Ĥ for the infinite square well), some operator has only continuous
spectrum (Ĥ for free particle), some has both (Ĥ for the finite square well)
8.2
8.2.1
Discrete Spectrum
Eigenvalues are Real
Suppose f is an eigenfunction of the operator Q̂, with eigenstate equation
Q̂f = qf
Do an inner product with f , in the inner product notation (which we will later find out to be a part of Dirac
notation)
D
E
f Q̂f = ⟨f |qf ⟩ = q ⟨f |f ⟩
15
Since Q̂ is Hermitian, we can move it to the front
E
D
E D
f Q̂f = Q̂f f = ⟨qf |f ⟩ = q ∗ ⟨f |f ⟩
Note that when the complex number q is at the front, since in the inner product the front part has to be taken
a complex conjugate, when we take it out, we take out its complex conjugate q ∗
Finally,
q ⟨f |f ⟩ = q ∗ ⟨f |f ⟩ =⇒ q = q ∗
8.2.2
Eigenfunctions Associated with Different Eigenvalues are Orthogonal
Suppose f is an eigenfunction of Q̂ associated to eigenvalue q, while g is an eigenfunction associated to q ′
We have the following eigenstate equation
(
Q̂f = qf
Q̂g = q ′ g
Do an inner product with g to the first equation to the back, and an inner product with f to the second equation
to the front
D
E
 f Q̂g = ⟨f |q ′ g⟩ = q ′ ⟨f |g⟩
E
D
 Q̂f g = ⟨qf |g⟩ = q ∗ ⟨f |g⟩
Since Q̂ is Hermitian we have
E
E D
D
f Q̂g = Q̂f g
Then,
q ′ ⟨f |g⟩ − q ∗ ⟨f |g⟩ = (q ′ − q ∗ ) ⟨f |g⟩ = 0
We have showed that eigenvalues of discrete spectrum are real, so q ∗ = q, while by assumption q ̸=′ . thus
⟨f |g⟩ = 0
8.2.3
Eigenfunctions are Complete
All wavefunction (which has to lie in Hilbert space by definition) can be expanded as a linear combination of
eigenfunctions
X
Ψ=
Cn ψn
(43)
n
where ψn are the eigenfunctions and Cn are the corresponding coefficient (this notation is consistent how we
represent the stationary state solution in Chapter 2, because stationary states are just the eigenfunctions of
Hamiltonian)
8.3
8.3.1
Continuous Spectrum
Eigenfunction of Momentum Operator
To solve for the eigenfunction, we write down the eigenstate equation
p̂fp = −iℏ
d
fp = pfp
dx
where p is the eigenvalue momentum
The general solution is
fp = Aeipx/ℏ
Now, we want to construct an orthonormal condition for the momentum operator, through which we will specify
the constant A
Do an inner product of two eigenfunctions associated with different eigenvalues p and p′
Z ∞
Z ∞
′
2
2
⟨fp′ |fp ⟩ =
fp∗′ (x)fp (x)dx = ∥A∥
e−ip x/ℏ eipx/ℏ dx = ∥A∥ 2πℏδ(p − p′ )
−∞
−∞
If we choose
r
A=
16
1
2πℏ
Then we get the following orthonormal condition, called Dirac orthonomality
⟨fp′ |fp ⟩ = δ(p − p′ )
(44)
We also have the completeness condition as follows: Any wave function f (x) can be represented as
Z ∞
Z ∞
1
c(p)eipx/ℏ dp
f (x) =
c(p)fp (x)dp = √
2πℏ −∞
−∞
(45)
The constant c(p) can be found by Fourier’s Trick (i.e. do an inner product with the orthogonal function)
Z ∞
Z ∞
⟨fp′ |f ⟩ =
c(p) ⟨fp′ |fp ⟩ dp =
c(p)δ(p − p′ )dp = c(p′ )
−∞
−∞
Although the inner product in this notation actually represent an integral, it satisfies the linearity of the inner
product of vectors, so we can take the integral and constant out of the braket.
8.3.2
Eigenfunction of the Position Operator
To solve for the eigenfunction, we write down the eigenstate equation
x̂gy (x) = ygy (z)
We know the position operator is just x (in the position basis), so
xgy (x) = ygy (x)
The only function that satisfies this condition is
gy (x) = Aδ(x − y)
Still, we want to construct a Dirac orthonomality for the position operator eigenfunction, similarly, we do an
inner product of two eigefunction associated with distinct eigenvalues
Z ∞
Z ∞
2
2
′
∗
gy gy =
gy′ (x)gy (x)dx = ∥A∥
δ(x − y ′ )δ(x − y)dx = ∥A∥ δ(y − y ′ )
−∞
−∞
Choose A = 1, we get the orthonormal condition
⟨gy′ |gy ⟩ = δ(y − y ′ )
The completeness condition is
(46)
∞
Z
f (x) =
c(y)gy (x)dy
(47)
−∞
9
9.1
Generalized Statistical Interpretation
Probability of Measurement
For discrete spectrum, the probability of getting a measurement result qn is
2
∥cn ∥ ,
where
cn = ⟨fn |Ψ⟩
(48)
For continuous spectrum, the probability of getting a measurement result in a range z + dz is
2
∥c(z)∥ dz,
9.2
where
c(z) = ⟨fz |Ψ⟩
(49)
Expectation Value
For discrete spectrum, suppose qn is the eigenvalues of the operator Q̂, where the coefficients of normalized
eigenfunctions are cn , then the expectationn value of Q̂ is
X
2
⟨Q⟩ =
qn ∥cn ∥
(50)
n
This can be shown by writing the function in expansion form
*
+ *
+
X
X
X
X
X
X
2
⟨Q⟩ =
cm ψm Q̂
cn ψn =
cm ψ m
cn qn ψn =
c∗m cn qn ⟨ψm |ψn ⟩ =
qn ∥cn ∥
m
n
m
n
Similarly, for continuous spectrum
Z
∞
⟨Q⟩ =
m,n
2
z∥c(z)∥ dz
−∞
17
n
(51)
9.3
Position and Momentum Space Wavefunction from Generalized Perspective
We can find the coefficient of eigenfunction according to equation (48), we can think of the position space and
momentum space wavefunction as such coefficient, which describes the contribution of each eigenvalue.
The position space wavefunction
Z
∞
c(y) = ⟨gy |Ψ⟩ =
δ(x − y)Ψ(x, t)dx = Ψ(y, t)
(52)
−∞
which is just the original wavefunction with variable relabelled. Note that this is because we are doing the
projection entirely in position basis, in momentum basis, we need to do the position operator to get the position
space wavefunction
The momentum space wavefunction
c(p) = ⟨fp |Ψ⟩ = √
1
2πℏ
Z
∞
e−ipx/ℏ Ψ(x, t)dx = Φ(p, t)
(53)
−∞
Another interpretation of equation is that the momentum space wavefunction is just the Fourier transformation
of the position space wavefunction
Similarly, the position space wavefunction is a inverse Fourier transformation of the momentum space wavefucntion
Z ∞
1
eipx/ℏ Φ(p, t)dp
(54)
Ψ(x, t) = √
2πℏ −∞
We can think of the momentum and position wavefunction as the projection of generalized wavefunction onto
the position or momentum basis, as we will show later
10
10.1
Uncertainty Principle
Generalized Uncertainty Principle
Theorem
For any observable A and B, the following relation about their uncertainty holds
Dh
iE2
1
2 2
σA σB ≥
Â, B̂
2i
Example
For position operator x̂ and p̂
[x̂, p̂] = iℏ
Thus,
σx2 σp2
2
ℏ
≥
2
This is the position-momentum uncertainty
Proof
Recall that we can express the uncertainty of an observable A in the following form
D
E D
E
2
σA
= Ψ (Â − ⟨A⟩)2 Ψ = (Â − ⟨A⟩)Ψ (Â − ⟨A⟩)Ψ
Thus, let
(
(Â − ⟨A⟩) = f
(B̂ − ⟨B⟩) = g
We can express the uncertainty of A and B in terms of these two functions
(
2
σA
= ⟨f |f ⟩
2
σB = ⟨g|g⟩
Then by Schwartz inequality (39), we have
2
2 2
σA
σB = ⟨f |f ⟩ ⟨g|g⟩ ≥ ∥⟨f |g⟩∥
18
(55)
Then just calculate the right side we will be able to obtain the uncertainty relation
E
D
⟨f |g⟩ = (Â − ⟨A⟩)Ψ (B̂ − ⟨B⟩)Ψ
Move the operator on the front to the back, expand everything, and use the linearity of inner product to separate
terms and move the constant out of the braket, we get
D
E
D
E
D
E
⟨f |g⟩ = Ψ ÂB̂Ψ − ⟨B⟩ Ψ ÂΨ − ⟨A⟩ Ψ B̂Ψ + ⟨A⟩ ⟨B⟩ ⟨Ψ|Ψ⟩
D
E
= ÂB̂ − ⟨B⟩ ⟨A⟩ − ⟨A⟩ ⟨B⟩ + ⟨A⟩ ⟨B⟩
D
E
= ÂB̂ − ⟨A⟩ ⟨B⟩
Similarly, we get (we can show this by relabeling the operator)
D
E
⟨g|f ⟩ = B̂Â − ⟨B⟩ ⟨A⟩
Then
D
E D
E Dh
iE
∗
⟨f |g⟩ − ⟨f |g⟩ = ⟨f |g⟩ − ⟨g|f ⟩ = ÂB̂ − B̂Â = Â, B̂
The last identity holds because the expectation value is just an inner product which has linearity Any complex
number has following property
2
1
2
∗
∥z∥ =
(z − z )
(56)
2i
Thus, since ⟨f |g⟩ is just a complex number
2
2 2
σA
σB = ⟨f |f ⟩ ⟨g|g⟩ ≥ ∥⟨f |g⟩∥ =
10.2
10.2.1
1
(⟨f |g⟩ − ⟨g|f ⟩)
2i
2
=
i2
1h
Â, B̂
2i
Energy Time Uncertainty Principle
Statement and Interpretation
The energy-time uncertainty principle is
ℏ
(57)
2
Here ∆E is the uncertainty in energy σH , ∆t is the time it takes for the system to change substantially, we will
later explain how this is defined exactly
∆E∆t ≥
10.2.2
Proof
Suppose we have an observable Q, take the time derivative of its expectation value
E
d
d D
⟨Q⟩ =
Ψ Q̂Ψ
dt
dt
+ *
∂ Q̂
dΨ
dΨ
=
Q̂Ψ + Ψ
Ψ + Ψ Q̂
dt
∂t
dt
We can replace all the time derivative of Ψ through the Schrodinger equation
1
dΨ
= ĤΨ
dt
iℏ
Then (note that in the first term, we we take the constant out, we take out the complex conjugate)
*
+
E
E
d
1 D
∂ Q̂
1 D
Ψ Q̂ĤΨ
⟨Q⟩ = −
ĤΨ Q̂Ψ + Ψ
Ψ +
dt
iℏ
∂t
iℏ
*
+
E
E
1 D
∂ Q̂
1 D
=−
Ψ ĤQ̂Ψ +
Ψ Q̂ĤΨ +
iℏ
iℏ
∂t
*
+
*
+
iE
iE
1 Dh
∂ Q̂
i Dh
∂ Q̂
=
Q̂, Ĥ +
=
Ĥ, Q̂ +
iℏ
∂t
ℏ
∂t
19
We get the last line by linearity
Assume that Q̂ does not depend on t explicitly (this is true for all the operators we encountered)
Dh
iE ℏ d ⟨Q⟩
Ĥ, Q̂ =
i dt
We have shown in the generalized uncertainty principle (55) that
2 2
σH
σQ
≥
iE2 1 ℏ d ⟨Q⟩ 2 ℏ 2 d ⟨Q⟩ 2
1 Dh
=
Ĥ, Q̂
=
2i
2i i dt
2
dt
Hence,
ℏ d ⟨Q⟩
2
dt
σH σQ ≥
Define
(
∆E = σH ∆t = σQ /
d⟨Q⟩
dt
(58)
Note that in the definition of t, we are comparing how fast that the expectation value of Q changes compared to
the uncertainty of Q, if the system changes very fast, then norm d⟨Q⟩
dt is large, and the time it takes the system
to change substantially ∆t will be very small
Finally, with the new definition introduced
ℏ
∆E∆t ≥
2
For different operator Q̂, the exact value of ∆E∆t can be different, but the uncertainty relation will hold for
any (physical) operator
11
11.1
Change of Basis
Generalized Wave Function
We define a generalized wavefunction |S(t)⟩, whose projection onto different basis is just the representation of
wavefunction in different basis
Projection on position basis
Ψ(x, t) = ⟨x|S⟩
(59)
where x is the eigenfunction of position operator x̂ with eigenvalue x
Projection on momentum basis
Φ(p, t) = ⟨p|S⟩
(60)
where p is the eigenfunction of position operator p̂ with eigenvalue p
Projection on energy basis
cn (t) = ⟨n|S⟩
(61)
where n is the eigenfunction of Hamiltonian operator with eigenvalue n
Note that all these projection gives a coefficient which describes the contribution of each state, we can sum (or
integral) of these projection (which is a function of the eigenvalue) and the corresponding eigenfucntion (which
is also a function of the eigenvalue) to get the generalized wavefunction
Z
|S(t)⟩ −→ Ψ(y, t)δ(x − y)dy
Z
1
eipx/ℏ dp
= Φ(p, t) √
(62)
2πℏ
X
=
cn ψn (x)e−iEn t/ℏ
n
where all these three expressions gives a function of x
The generalized wavefunction can be analogously think of as a vector, it does not have a specific form unless a
basis is specified
20
11.2
Wavefunction as Vectors
Suppose |en ⟩ is a orthonormal basis, then a vector |α⟩ can be expanded on it
X
|α⟩ =
an |en ⟩
n
where the coefficient
an = ⟨en |α⟩
11.3
Operator as Matrix
An operator transforms a vector
|β⟩ = Q̂ |α⟩
th
The m, n
entry of the matrix corresponds to Q̂ is defined by
Qmn = ⟨em | Q̂ |en ⟩
(63)
The suppose we have two vector expanded on the orthonormal basis
(
P
|α⟩ = n an |en ⟩
P
|β⟩ = n bn |en ⟩
To find the component of |β⟩ from the components of α, we can do an inner product with ⟨em |
X
X
bn |en ⟩ =
an Q̂ |en ⟩
n
X
n
bn ⟨em |en ⟩ =
n
X
X
bn δmn = bm =
n
11.4
an ⟨em | Q̂ |en ⟩
n
X
an Qmn
n
Projection Operator
Suppose |α⟩ is a normalized vector, then the projection operator
P̂ = |α⟩ ⟨α|
It picks out the component of any vector along |α⟩
11.5
Identity Operator
If |en ⟩ is a discrete orthonormal basis, then
⟨em |en ⟩ = 1
The identity operator is
X
|en ⟩ ⟨en | = 1
n
If |ez ⟩ is a Dirac orthonormal basis, then
⟨ez′ |ez ⟩ = δ(z − z ′ )
The identity operator is
Z
|ez ⟩ ⟨ez | dz = 1
21
(64)
11.6
Change of Basis
By our previous discussion on the identity operators in position, momentum, and energy basis are respectively:
R

R |x⟩ ⟨x| dx
(65)
|p⟩ ⟨p| dp
P

n |n⟩ ⟨n|
This notation allow us to easily project the abstract wavefunction onto different basis, for example
Z
Z
|S(t)⟩ = dx |x⟩ ⟨x|S(t)⟩ = Ψ(x, t) |x⟩ dx
Z
Z
|S(t)⟩ = dp |p⟩ ⟨p|S(t)⟩ = Φ(p, t) |p⟩ dp
X
X
|S(t)⟩ =
|n⟩ ⟨n|S(t)⟩ =
cn (t) |n⟩
n
(66)
n
This is just an equivalent expression for equation (62
We can also represent operators in different basis, for example
⟨x| x̂ |S⟩
Is the action of the position operator on the wavefunction represented in position basis
⟨p| x̂ |S⟩
Is the action of position operator
R in the momentum space, we will be able to determine its specific form by
inserting the identity operator |x⟩ ⟨x| dx before |S⟩
22