Name_______________________________
Dirac Notation Homework
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Throughout this homework, the normalization issues for the position and momentum eigenstates are ignored.
̂.
In all of the questions below, |Ψ⟩ denotes a generic state of a quantum system with Hamiltonian 𝐻
Assume that the Hilbert space is infinite dimensional.
For all questions involving a generic operator 𝑄̂ , assume that it only depends on position 𝑥̂ and momentum 𝑝̂ and has no
explicit time dependence.
Assume the particle is confined in a one dimensional physical “laboratory” space.
Assume ∑𝑛 refers to a summation over a complete set of states (𝑛 = 1,2,3 … ∞).
The goals of this homework are to help you learn about:
 Connecting Dirac notation with function space (position and momentum representation)
 Relationship between state vector |𝚿⟩ and the wavefunction 𝚿(𝒙)
o In the position representation, ⟨𝑥|Ψ⟩ = Ψ(𝑥) is the projection of state vector |Ψ⟩ along the
eigenstate of position |𝑥⟩ (position eigenstates {|𝑥⟩} form a complete set of basis vectors). The
column vector Ψ(𝑥) (considered as a function of 𝑥) is called the position space wavefunction.
o In the momentum representation, ⟨𝑝|Ψ⟩ = Φ(𝑝) is the projection of state vector |Ψ⟩ along the
eigenstate of momentum |𝑝⟩ (momentum eigenstates {|𝑝⟩} form a complete set of basis vectors).
The column vector Φ(𝑝) (considered as a function of 𝑝) is called the momentum space
wavefunction.
o The position space wavefunction Ψ(𝑥) and momentum space wavefunction Φ(𝑝) can be
considered as infinite dimensional column vectors.
o The momentum space wavefunction Φ(𝑝) = ⟨𝑝|Ψ⟩ and position space wavefunction Ψ(𝑥) =
⟨𝑥|Ψ⟩ are a Fourier transform and inverse Fourier transform of each other, respectively.
 Position and momentum eigenstates
o The position eigenstate with eigenvalue 𝑥 ′ can be written as:
 |𝑥′⟩ in Dirac notation.
 ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ) in position representation. It is also called the position eigenfunction
in position representation and a form of the orthogonality relation.
−𝑖𝑝𝑥′

o
⟨𝑝|𝑥′⟩ =
𝑒 ℏ
⟨𝑥|𝑝′⟩ =
𝑒 ℏ
in the momentum representation. It is also called the position
eigenfunction in momentum representation.
The momentum eigenstate with eigenvalue 𝑝′ can be written as:
 |𝑝′⟩ in Dirac notation
√2𝜋ℏ
𝑖𝑝′ 𝑥


in the position representation.
It is also called the momentum
eigenfunction in position representation.
 ⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ ) in the momentum representation. It is also called the momentum
eigenfunction in momentum representation and a form of the orthogonality relation.
Position and momentum representation
o For a generic state |Ψ⟩ and a generic operator 𝑄̂ (which depends only on operators 𝑥̂ and 𝑝̂ ):
𝜕
 In the position representation, 𝑄̂ |Ψ⟩ is represented by ⟨𝑥|𝑄̂|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥)
√2𝜋ℏ
𝜕
In the momentum representation, 𝑄̂ |Ψ⟩ is represented by ⟨𝑝|𝑄̂|Ψ⟩ = 𝑄̂ (𝑖ℏ 𝜕𝑝 , 𝑝) Φ(𝑝)
̂ and its eigenstates (energy eigenstates)
 For a given quantum system, the Hamiltonian operator 𝑯

o
o
o
|𝐸𝑛 ⟩ = |Ψ𝐸𝑛 ⟩ = |Ψ𝑛 ⟩ for 𝑛 = 1,2, … ∞, represent a complete set of energy eigenstates with
eigenvalue 𝐸𝑛 .
̂ |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩.
The eigenvalue equation for the Hamiltonian operator is 𝐻
̂
The expectation value of energy in state |Ψ⟩ is ⟨Ψ|𝐻 |Ψ⟩ = ∑𝑛 𝐸𝑛 |⟨𝜓𝑛 |Ψ⟩|2 .
Connecting Dirac notation with function space (position and momentum representation):
 Relationship between state vector |𝚿⟩ and the wavefunction for a given quantum system
1. Choose all of the following statements that are correct about the generic state vector |Ψ⟩
(I)
Ψ(𝑥) = ⟨𝑥|Ψ⟩ is a representation of the state vector |Ψ⟩ in the position
representation and is called the position space wavefunction.
(II)
Φ(𝑝) = ⟨𝑝|Ψ⟩ is a representation of the state vector |Ψ⟩ in the momentum
representation and is called the momentum space wavefunction.
(III)
The state vector |Ψ⟩ can be written as a column vector once a basis has been chosen.
(a)
(I) only
(b)
(I) and (II) only
(c)
(I) and (III) only
(d)
(II) and (III) only
(e)
All of the above.
2. Consider the following conversation between two students:
 Student A: In the preceding question, we can also write Ψ(𝑥) as |Ψ⟩ = Ψ(𝑥).
 Student B: I disagree. |Ψ⟩ ≐ Ψ(𝑥), where the ≐ sign means that the equality is valid only with
respect to a chosen basis. Ψ(𝑥) is a representation of |Ψ⟩ in position representation when we choose
a complete set of position eigenstates, |𝑥⟩, as our basis vectors, so we can write Ψ(𝑥) = ⟨𝑥|Ψ⟩.
With whom do you agree? Explain your reasoning.
3. Earlier you learned that in a three dimensional space, a generic state |𝐺⟩ can be written as |𝐺⟩ = 𝑎|𝑖⟩ + 𝑏|𝑗⟩ +
𝑐|𝑘⟩, where |𝑖⟩, |𝑗⟩, and |𝑘⟩ form an orthonormal basis and 𝑎, 𝑏, and 𝑐 are complex numbers. Thus, a generic
state |𝐺⟩ can be written as a column vector once a basis is chosen, i.e.,
⟨𝑖|𝐺⟩
𝑎
|𝐺⟩ ≐ ( ⟨𝑗|𝐺⟩ ) = (𝑏).
𝑐
⟨𝑘|𝐺⟩
where the ≐ sign means that the equality is valid only with respect to a chosen basis.
In quantum mechanics, the generic state vector |Ψ⟩ is a vector in the Hilbert space, which is infinite
dimensional. The generic state |Ψ⟩ can be written in terms of a linear superposition of a complete set of
eigenstates of any hermitian operator 𝑄̂ .
(a)
Suppose a hermitian operator 𝑄̂ has a complete set of eigenstates {|𝑞𝑛 ⟩, 𝑛 = 1,2, … ∞} with
discrete eigenvalues 𝑞𝑛 . Write |Ψ⟩ in terms of a linear superposition of a complete set of
eigenstates {|𝑞𝑛 ⟩, 𝑛 = 1, 2, 3 … ∞}.
2
(b)
Keeping in mind how the generic state |𝐺⟩ is written as a column vector, write |Ψ⟩ as an infinite
dimensional column vector with respect to the orthonormal basis |𝑞𝑛 ⟩.
4. Consider a generic state vector |Ψ⟩ = ∑∞
𝑛=1 𝑎𝑛 |𝑞𝑛 ⟩, where 𝑎𝑛 = ⟨𝑞𝑛 |Ψ⟩. Then, the state vector |Ψ⟩ can be
𝑎1
⟨𝑞1 |Ψ⟩
represented as a column vector like this: |Ψ⟩ ≐ (⟨𝑞2 |Ψ⟩) = (𝑎2 ), where the ≐ sign means that this is a
⋮
⋮
|Ψ⟩
⟩,
representation of
in the chosen basis, e.g., {|𝑞𝑛 𝑛 = 1,2, … ∞}. Consider the following conversation
between two students about the situation where basis vectors are chosen to be position eigenstates |𝑥⟩ or
momentum eigenstates |𝑝⟩, each of which have a continuous eigenvalue spectrum.
 Student 1: We cannot write state vector |Ψ⟩ as a column vector if position eigenstates |𝑥⟩ are chosen
as the basis vectors. State vector |Ψ⟩ written as a linear superposition of position eigenstates |𝑥⟩ is
∞
|Ψ⟩ = ∫−∞ 𝐶𝑥 |𝑥⟩𝑑𝑥, where 𝐶𝑥 = ⟨𝑥|Ψ⟩. But since this expansion of |Ψ⟩ involves an integral instead
of a summation, we cannot write |Ψ⟩ as a column vector with respect to the basis vectors |𝑥⟩.
 Student 2: I disagree. Even though the expansion of |Ψ⟩ is an integral instead of a summation, we
can still envision |Ψ⟩ as a column vector with respect to the basis vectors |𝑥⟩. Like this:
𝑐1
⟨𝑥1 |Ψ⟩
Ψ(𝑥1 )
|Ψ⟩ ≐ (𝑐2 ) = (⟨𝑥2 |Ψ⟩) = (Ψ(𝑥2 )) = Ψ(𝑥).
⋮
⋮
⋮
 Student 1: But why do the 𝑥’s have indices? Don’t position eigenstates |𝑥⟩ have a continuous
eigenvalue spectrum 𝑥, not a discrete eigenvalue spectrum?
 Student 2: Yes, you are correct. Actually, you should think of 𝑥1 = ∆𝑥, 𝑥2 = 2∆𝑥 … etc., and take
the limit as ∆𝑥 → 0. I was simply making an analogy with the discrete eigenvalue spectrum case.
However, the best way to write |Ψ⟩ when position eigenstates |𝑥⟩ are chosen as the basis vectors is as
Ψ(𝑥), which is also called the position space wavefunction. Ψ(𝑥) is a column vector with position
eigenvalues 𝑥 as a continuous index.
Do you agree with Student 2? Explain your reasoning.
5. Consider the following conversation between three students:
 Student A: How does this expansion of |Ψ⟩ in terms of a complete set of position eigenstates |𝑥⟩,
∞
|Ψ⟩ = ∫−∞ 𝐶𝑥 |𝑥⟩𝑑𝑥, help you in questions involving the measurement of the position of the particle?




2
Student B: |𝐶𝑥0 | 𝑑𝑥 = |⟨𝑥0 |Ψ⟩|2 𝑑𝑥 = |Ψ(𝑥0 )|2 𝑑𝑥 gives us the probability of finding the particle in
a narrow range between 𝑥0 and 𝑥0 + 𝑑𝑥 when we measure the position of the particle.
Student C: But I thought that the probability of finding the particle in a narrow range between 𝑥0 and
𝑥 +𝑑𝑥
𝑥0 + 𝑑𝑥 was ∫𝑥 𝑜
𝑥|Ψ(𝑥)|2 𝑑𝑥 .
0
Student A: So is the probability of finding the particle in a narrow range between 𝑥0 and 𝑥0 + 𝑑𝑥
𝑥 +𝑑𝑥
represented mathematically as ∫𝑥 𝑜
𝑥|Ψ(𝑥)|2 𝑑𝑥 = 𝑥0 |Ψ(𝑥0 )|2 𝑑𝑥?
0
Student B: No. It is just |Ψ(𝑥0 )|2 𝑑𝑥 not 𝑥0 |Ψ(𝑥0 )|2 𝑑𝑥. |Ψ(𝑥0 )|2 is the probability density at
position 𝑥0 . We multiply |Ψ(𝑥0 )|2 by a width 𝑑𝑥 to obtain the probability of finding the particle in a
narrow range between 𝑥0 and 𝑥0 + 𝑑𝑥.
3
Do you agree with Student B’s explanation? Explain your reasoning.
6. Write the generic state vector |Ψ⟩ as a column vector in the momentum representation when momentum
eigenstates |𝑝⟩ are chosen as basis vectors.
7. Write the probability of finding the particle with a momentum between 𝑝0 and 𝑝0 + 𝑑𝑝 when we measure the
momentum of the particle.
Position and Momentum Eigenstates
 Position eigenstates
7. Choose all of the following equations that are correct for the position eigenstate |𝑥′⟩ with eigenvalue 𝑥 ′ .
(I)
𝑥̂|𝑥′⟩ = 𝑥′|𝑥′⟩
⟨𝑥|𝑥̂|𝑥′⟩ = 𝑥′⟨𝑥|𝑥 ′ ⟩ = 𝑥′𝛿(𝑥 − 𝑥 ′ )
(II)
⟨𝑥|𝑥̂|𝑥′⟩ = 𝑥𝛿(𝑥 − 𝑥 ′ ) = 𝑥′𝛿(𝑥 − 𝑥 ′ )
(III)
(a)
(I) and (II) only
(b)
(I) and (III) only
(c)
(II) and (III) only
(d)
All of the above.
8. Consider the following conversation between Student A and Student B.
 Student B: I don’t understand how 𝑥̂|𝑥′⟩ = 𝑥′|𝑥′⟩ and ⟨𝑥|𝑥̂|𝑥′⟩ = 𝑥𝛿(𝑥 − 𝑥 ′ ) = 𝑥′𝛿(𝑥 − 𝑥 ′ ) can both be
correct. They are two different eigenvalue equations for the same operator 𝑥̂, so shouldn’t one of them be
incorrect?
 Student A: Actually, 𝑥̂|𝑥′⟩ = 𝑥′|𝑥′⟩ and 𝑥𝛿(𝑥 − 𝑥 ′ ) = 𝑥′𝛿(𝑥 − 𝑥 ′ ) convey the same information. The
former is the eigenvalue equation for the position operator 𝑥̂ in Dirac notation, and the latter is the
eigenvalue equation for the position operator in the position representation. In the position representation,
𝑥̂ is equivalent to a multiplication by 𝑥. We can also write ⟨𝑥|𝑥̂|𝑥′⟩ = 𝑥𝛿(𝑥 − 𝑥 ′ ) = 𝑥 ′ 𝛿(𝑥 − 𝑥 ′ ) since
⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ), which is zero for all position eigenvalues except when 𝑥 = 𝑥 ′ .
With whom do you agree? Explain.
4
9. A) Which one of the following is the eigenstate of position |𝑥′⟩ with eigenvalue 𝑥 ′ written in position
representation, i.e., ⟨𝑥|𝑥′⟩?
⟨𝑥|𝑥′⟩ = Ψ(𝑥 ′ )
(a)
⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ )
(b)
𝑖𝑝𝑥′
𝑒 ℏ
(c)
⟨𝑥|𝑥′⟩ =
(d)
⟨𝑥|𝑥′⟩ = Ψ(𝑥)
√2𝜋ℏ
9. B) The position eigenstate written in position representation (when considered as a function of 𝑥) is called
the:
(a)
Position eigenfunction in position representation.
(b)
Position eigenfunction in momentum representation.
(c)
Position eigenfunction either in position or momentum representation since the expression for
position eigenfunction is the same regardless of the representation.
(d)
None of the above.
10. Consider the following conversation between two students:
 Student 1: The position eigenfunction should always be a delta function whether we write it in
position or momentum representation.
 Student 2: I disagree. When position eigenstate |𝑥′⟩ with eigenvalue 𝑥 ′ is written by choosing
position eigenstates |𝑥⟩ as basis vectors, we obtain ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ) which is the position
eigenfunction in the position representation. When the position eigenstate |𝑥′⟩ is written by choosing
momentum eigenstates |𝑝⟩ as basis vectors, we obtain ⟨𝑝|𝑥′⟩, which is the position eigenfunction in
the momentum representation. However, ⟨𝑝|𝑥′⟩ is not a delta function. The position eigenfunction is
only a delta function in the position representation, but not in the momentum representation.
With whom do you agree? Explain your reasoning.
11. Which one of the following is the eigenstate of position |𝑥⟩ with eigenvalue 𝑥 written in momentum
representation, i.e., ⟨𝑝|𝑥⟩ (position eigenfunction in momentum representation)?
−𝑖𝑝𝑥
𝑒 ℏ
(a)
⟨𝑝|𝑥⟩ = Ψ𝑥 (𝑝) =
(b)
(c)
(d)
⟨𝑝|𝑥⟩ = Ψ𝑥 (𝑝) = 𝛿(𝑝 − 𝑥)
⟨𝑝|𝑥⟩ = Ψ𝑥 (𝑝) = 𝛿(𝑝 − 𝑝′)
⟨𝑝|𝑥⟩ = Ψ𝑥 (𝑝) = 𝛿(𝑥 − 𝑥′)
√2𝜋ℏ
5
 Momentum eigenstates
12. Choose all of the following equations that are correct for the momentum eigenstate |𝑝′⟩ with eigenvalue 𝑝′ .
(I)
𝑝̂ |𝑝′⟩ = 𝑝′|𝑝′⟩
⟨𝑝|𝑝̂ |𝑝′⟩ = 𝑝′⟨𝑝|𝑝′ ⟩ = 𝑝′𝛿(𝑝 − 𝑝′ )
(II)
⟨𝑝|𝑝̂ |𝑝′⟩ = 𝑝𝛿(𝑝 − 𝑝′ ) = 𝑝′ 𝛿(𝑝 − 𝑝′ )
(III)
(a)
(I) and (II) only
(b)
(I) and (III) only
(c)
(II) and (III) only
(d)
All of the above.
13. Which one of the following is the eigenstate of momentum |𝑝′⟩ with eigenvalue 𝑝′ in momentum
representation, i.e., ⟨𝑝|𝑝′⟩ (momentum eigenfunction in momentum representation)?
⟨𝑝|𝑝′⟩ = Φ(𝑝′ )
(a)
⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ )
(b)
𝑖𝑝′𝑥
(c)
⟨𝑝|𝑝′⟩ =
𝑒 ℏ
√2𝜋ℏ
−𝑖𝑝′𝑥
(d)
⟨𝑝|𝑝′⟩ =
𝑒 ℏ
√2𝜋ℏ
14. Which one of the following is the eigenstate of momentum in position representation, i.e., ⟨𝑥|𝑝⟩ (momentum
eigenfunction in position representation)?
⟨𝑥|𝑝⟩ = Ψ𝑝 (𝑥) = 𝛿(𝑥 − 𝑥 ′ )
(a)
𝑖𝑝𝑥
𝑒 ℏ
(b)
⟨𝑥|𝑝⟩ = Ψ𝑝 (𝑥) =
(c)
(d)
⟨𝑥|𝑝⟩ = Ψ𝑝 (𝑥) = 𝛿(𝑝 − 𝑝′ )
⟨𝑥|𝑝⟩ = Ψ𝑝 (𝑥) = 𝛿(𝑥 − 𝑝)
√2𝜋ℏ
15. Substitute the expression you chose for the momentum eigenfunction in the position representation in the
preceding question and the expression for the momentum operator (one dimensional) in position
𝜕
representation, 𝑝̂ = −𝑖ℏ 𝜕𝑥 , in the eigenvalue equation for momentum, 𝑝̂ Ψ𝑝′ (𝑥) = 𝑝′Ψ𝑝′ (𝑥), where Ψ𝑝′ (𝑥) is
the eigenstate of momentum with eigenvalue 𝑝’ in position representation Check whether the eigenvalue
equation is satisfied. If not, correct your answer to the preceding question. Show all your work in the space
provided below.
6
16. The eigenstates of a hermitian operator with different eigenvalues are orthonormal1 to each other. Position
and momentum operators (𝑥̂ and 𝑝̂ ) each have a continuous eigenvalue spectrum. Assume that the
̂ ) for a given quantum system has a discrete eigenvalue spectrum. Choose all of the
Hamiltonian operator (𝐻
following that are correct about the scalar product of two eigenstates of an operator.
⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ )
(I)
⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ )
(II)
⟨𝐸|𝐸′⟩ = 𝛿𝐸𝐸′ where E denotes energy.
(III)
Note: In more common notation ⟨𝐸|𝐸′⟩ = 𝛿𝐸𝐸 ′ can be written as ⟨Ψ𝐸1 |Ψ𝐸2 ⟩ = ⟨Ψ1 |Ψ2⟩ = 𝛿12 ,
where |𝐸1 ⟩ = |Ψ𝐸1 ⟩ = |Ψ1 ⟩.
(a)
(b)
(c)
(d)
(III) only
(I) and (II) only
(I) and (III) only
All of the above.
17. Consider the following conversation between two students about a position eigenfunction and a momentum
eigenfunction in the position representation:
 Student 1: A position eigenstate is represented in position representation as ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ).
𝛿(𝑥 − 𝑥 ′ ) is a special type of position space wavefunction Ψ(𝑥) = ⟨𝑥|Ψ⟩ in which position has a definite
value of 𝑥 ′ . Before taking the limit to obtain the delta function, a position eigenfunction in the position
representation, 𝛿(𝑥 − 𝑥 ′ ), is a very sharply peaked wavefunction about 𝑥 = 𝑥 ′ in position representation.
 Student 2: I agree with you. In addition, a momentum eigenstate is represented in position representation
𝑖𝑝′ 𝑥
𝑒 ℏ
𝑖𝑝′ 𝑥
𝑒 ℏ
as ⟨𝑥|𝑝′⟩ = Ψ𝑝′ (𝑥) =
.
is another special type of position space wavefunction Ψ(𝑥) = ⟨𝑥|Ψ⟩
√2𝜋ℏ √2𝜋ℏ
where momentum has a definite value 𝑝′ . Instead of being sharply peaked like a delta function, a
𝑖𝑝′ 𝑥
𝑒 ℏ
momentum eigenfunction in the position representation is spread out as
(which is a linear
√2𝜋ℏ
combination of sine and cosine functions over all position with a definite momentum 𝑝’ and wave number
𝑝′
𝑘’ = ) and the probability density is uniform.
ℏ
Do you agree with Student 1, Student 2, both, or neither? Explain your reasoning.
|𝑥′⟩ and |𝑥⟩ or |𝑝′⟩ and |𝑝⟩ are not actually normalized because the inner products are ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ) and
⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ ). 𝛿(𝑥 − 𝑥 ′ ) and 𝛿(𝑝 − 𝑝′ ) are not normalized because they diverge when 𝑥 = 𝑥 ′ or 𝑝 = 𝑝′ ,
respectively. It is actually the integral over all space of a Dirac delta function that equals 1, e.g.,
∞
∫−∞ 𝛿(𝑥 − 𝑥 ′ )𝑑𝑥 = 1 . However, we will ignore these normalization issues for position and momentum
eigenstates.
7
1
18. Consider the following conversation between two students:
 Student 1: ⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ ) is a special type of momentum space wavefunction Φ(𝑝) = ⟨𝑝|Ψ⟩, in
which momentum has a definite value 𝑝′ . A momentum eigenfunction in the momentum
representation, 𝛿(𝑝 − 𝑝′ ), is a very sharply peaked wavefunction about 𝑝 = 𝑝′ in momentum
representation.
 Student 2: I agree with you. In addition, a position eigenstate |𝑥′⟩ with eigenvalue 𝑥 ′ represented in
−𝑖𝑝𝑥′
momentum representation is ⟨𝑝|𝑥′⟩ = Φ𝑥 ′ (𝑝) =
𝑒 ℏ
√2𝜋ℏ
−𝑖𝑝𝑥′
.
𝑒 ℏ
√2𝜋ℏ
is also a special type of momentum
−𝑖𝑝𝑥′
′
′
space wavefunction Φ(𝑝) = ⟨𝑝|Ψ⟩ in which position has a definite value 𝑥 (𝑥 is fixed in
𝑒 ℏ
√2𝜋ℏ
,
which is a function of 𝑝).
Do you agree with Student 1, Student 2, neither, or both? Explain your reasoning.
19. Is the position eigenstate (or position eigenfunction) in the momentum representation that you learned about
in the preceding question localized or an extended function of momentum 𝑝 (consider the real and imaginary
parts)? Explain your reasoning.
20. You have learned about position and momentum eigenstates in the position representation and momentum
representation. Using what you have learned, fill out the table below:
Position representation
Momentum representation
Position eigenstates
(or position eigenfunctions)
Momentum eigenstates
(or momentum eigenfunctions)
8
 Position and Momentum Operators
Up to this point, you have learned the following eigenvalue equations:


𝑥̂|𝑥′⟩ = 𝑥′|𝑥′⟩ is an eigenvalue equation for position operator 𝑥̂ with eigenvalue 𝑥’ in Dirac notation.
o ⟨𝑥|𝑥̂|𝑥′⟩ = 𝑥𝛿(𝑥 − 𝑥 ′ ) = 𝑥′𝛿(𝑥 − 𝑥 ′ ) is an eigenvalue equation for position operator 𝑥̂ with
eigenvalue 𝑥 ′ in position representation.
o ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ) is the position eigenfunction with eigenvalue 𝑥 ′ in position representation and
the position operator in position representation is 𝑥̂ = 𝑥 (multiplication of the position space
wavefunction Ψ(𝑥) = 𝛿(𝑥 − 𝑥 ′ ) by 𝑥).
𝑝̂ |𝑝′⟩ = 𝑝′|𝑝′⟩ is the eigenvalue equation for momentum operator 𝑝̂ with eigenvalue 𝑝′ in Dirac notation.
o ⟨𝑝|𝑝̂ |𝑝′⟩ = 𝑝𝛿(𝑝 − 𝑝′ ) = 𝑝′𝛿(𝑝 − 𝑝′ ) is the eigenvalue equation for momentum operator 𝑝̂ with
eigenvalue 𝑝’ in momentum representation.
o ⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ ) is the momentum eigenfunction with eigenvalue 𝑝′ in momentum
representation and the momentum operator in momentum representation is 𝑝̂ = 𝑝 (multiplication
of the momentum space wavefunction Φ(𝑝) = 𝛿(𝑝 − 𝑝′ ) by 𝑝).
𝑖𝑝′ 𝑥
𝑒 ℏ
o
⟨𝑥|𝑝′⟩ = Ψ𝑝′ (𝑥) =
o
representation and the momentum operator in position representation is 𝑝̂ = −𝑖ℏ 𝜕𝑥.
⟨𝑥|𝑝̂ |𝑝′⟩ = 𝑝̂ Ψ𝑝′ (𝑥) = 𝑝′ Ψ𝑝′ (𝑥) is the eigenvalue equation for momentum operator 𝑝̂ with
eigenvalue 𝑝′ in position representation.
√2𝜋ℏ
is the eigenfunction of momentum with eigenvalue 𝑝′ in position
𝜕
21. You have not yet learned about the position operator in momentum representation. Which one of the
following equations would help you identify the position operator 𝑥̂ in momentum representation?
(a)
𝑥̂Ψ(𝑥) = 𝑥Ψ(𝑥), where Ψ(𝑥) = ⟨𝑥|Ψ⟩ is a generic state |Ψ⟩ written in position representation.
(b)
𝑥̂Φ𝑥 ′ (𝑝) = 𝑥 ′ Φ𝑥 ′ (𝑝) , where Φ𝑥 ′ (𝑝) = ⟨𝑝|𝑥′⟩ is the position eigenfunction with eigenvalue 𝑥 ′ in
momentum representation.
(c)
𝑥̂Φ(𝑝) = 𝑥Φ(𝑝), where Φ(𝑝) = ⟨𝑝|Ψ⟩ is a generic state |Ψ⟩ written in momentum representation.
(d)
𝑥̂Ψ𝑝′ (𝑥) = 𝑥 ′ Ψ𝑝′ (𝑥), where Ψ𝑝′ (𝑥) = ⟨𝑥|𝑝′⟩ is the momentum eigenfunction with eigenvalue 𝑝′ in
position representation.
22. The correct answer to the preceding question is (b). Consider the eigenvalue equation for the position
−𝑖𝑝𝑥′
′
𝑒 ℏ
operator in momentum representation, 𝑥̂Φ𝑥 ′ (𝑝) = 𝑥 Φ𝑥 ′ (𝑝) , in which Φ𝑥 ′ (𝑝) = ⟨𝑝|𝑥′⟩ =
is the
√2𝜋ℏ
′
position eigenfunction with eigenvalue 𝑥 in momentum representation. Which one of the following must be
the position operator in momentum representation to satisfy the eigenvalue equation 𝑥̂Φ𝑥 ′ (𝑝) = 𝑥 ′ Φ𝑥′ (𝑝)?
(a)
𝑥
𝜕
(b)
𝑖ℏ 𝜕𝑥
(c)
𝑖ℏ
(d)
𝑝
𝜕
𝜕𝑝
9
23. Substitute in the answer you selected in the preceding question (for the position operator in momentum
representation) into the eigenvalue equation for the position operator in momentum representation,
−𝑖𝑝𝑥′
′
𝑒 ℏ
𝑥̂Φ𝑥 ′ (𝑝) = 𝑥 Φ𝑥′ (𝑝), in which Φ𝑥 ′ (𝑝) = ⟨𝑝|𝑥′⟩ =
is the position eigenfunction with eigenvalue 𝑥 ′ in
√2𝜋ℏ
momentum representation. Check whether the eigenvalue equation is satisfied. If not, correct your answer to
the preceding question. Show all your work in the space provided below.
You now know the position operator 𝑥̂ and momentum operator 𝑝̂ in both the position and momentum
representations. Fill in the table below with your results:
Position Representation
Momentum Representation
Position operator 𝑥̂
Momentum operator 𝑝̂
10
24. Consider the following conversation between two students:
 Student 1: The position operator 𝑥̂ acting on a generic state |Ψ⟩ written in Dirac notation without
reference to a basis is 𝑥̂|Ψ⟩. Suppose we choose a basis in which the eigenstates of position, |𝑥⟩, are
chosen as basis vectors. 𝑥̂|Ψ⟩ is represented in position representation by taking the scalar product of
𝑥̂|Ψ⟩ with |𝑥⟩, like this: ⟨𝑥|𝑥̂|Ψ⟩ = 𝑥Ψ(𝑥).
 Student 2: I don’t see how that is correct. How did the state vector |Ψ⟩ turn into the position space
wavefunction Ψ(𝑥)?
∞
 Student 1: We can insert the identity operator written in terms of position eigenstates ∫−∞|𝑥′⟩⟨𝑥′|𝑑𝑥′
into the expression ⟨𝑥|𝑥̂|Ψ⟩, like this:
∞
∞
∞
∫−∞⟨𝑥|𝑥̂|𝑥′⟩⟨𝑥′|Ψ⟩𝑑𝑥′ = ∫−∞ 𝑥′⟨𝑥|𝑥′⟩⟨𝑥′|Ψ⟩𝑑𝑥′ = ∫−∞ 𝑥 ′ 𝛿(𝑥 − 𝑥 ′ )Ψ(𝑥 ′ )𝑑𝑥 ′ = 𝑥Ψ(𝑥).
 Student 2: I see. So the position operator 𝑥̂ acting on a generic wavefunction Ψ(𝑥) in position
representation just amounts to multiplication of Ψ(𝑥) by 𝑥.
Do you agree with Student 1’s explanation and Student 2’s statement? Explain your reasoning.
25. Choose all of the following that correctly describe the momentum operator 𝑝̂ in position representation acting
𝜕
on a generic position space wavefunction Ψ(𝑥), i.e., 𝑝̂ (𝑥, −𝑖ℏ 𝜕𝑥)Ψ(𝑥) = ?
(I)
𝑝̂ Ψ(𝑥) = 𝑝Ψ(𝑥)
𝜕
(II)
𝑝̂ Ψ(𝑥) = −𝑖ℏ 𝜕𝑥 Ψ(𝑥)
(III)
𝑝̂ Ψ(𝑥) = ⟨𝑥|𝑝̂ |Ψ⟩
(IV)
𝑝̂ Ψ(𝑥) = ⟨𝑝|𝑝̂ |Ψ⟩
(a)
(I) only
(b)
(II) only
(c)
(I) and (IV) only
(d)
(I), (II), and (III) only
11
26. Consider the following conversation between Student A and Student B:
 Student A: The eigenvalue equation for momentum 𝑝̂ |𝑝′⟩ = 𝑝′|𝑝′⟩ in position representation when
the basis vectors are chosen to be the eigenstates of position, |𝑥⟩, is given by the expression
𝑖𝑝′𝑥
⟨𝑥|𝑝̂ |𝑝′⟩ = 𝑝′ ⟨𝑥|𝑝′⟩ = 𝑝′ (
𝑒 ℏ
). This is because ⟨𝑥|𝑝̂ |𝑝′⟩ = ⟨𝑥|𝑝′|𝑝′⟩, and since 𝑝′ sandwiched in
√2𝜋ℏ
𝑖𝑝′ 𝑥
the middle is a number, we are left with 𝑝

′ ⟨𝑥|𝑝′⟩
′
=𝑝 (
𝑒 ℏ
).
√2𝜋ℏ
Student B: I agree with you. But we can also think about it as
𝑖𝑝′𝑥
′⟩
⟨𝑥|𝑝̂ |𝑝′⟩ = 𝑝̂ (𝑥)⟨𝑥|𝑝 =
𝜕
−𝑖ℏ 𝜕𝑥 ⟨𝑥|𝑝′⟩
=
𝜕 𝑒 ℏ
−𝑖ℏ 𝜕𝑥 (
)
√2𝜋ℏ
𝑖𝑝′𝑥
= 𝑝′ (
𝑒 ℏ
𝜕
), where 𝑝̂ = −𝑖ℏ 𝜕𝑥
√2𝜋ℏ
is the
𝑖𝑝′𝑥
momentum operator in position representation.
Here, ⟨𝑥|𝑝′⟩ = Ψ𝑝′ (𝑥) =
𝑒 ℏ
√2𝜋ℏ
is a momentum
eigenfunction with eigenvalue 𝑝′ in position representation (which is a special type of position space
wavefunction Ψ(𝑥) = ⟨𝑥|Ψ⟩, where |Ψ⟩ = |𝑝′⟩, a momentum eigenstate).
Do you agree with Student A and Student B? Explain why or why not.
Note: For a hermitian operator 𝑄̂ , the notation ⟨Ψ|𝑄̂ |Ψ⟩ with 𝑄̂ between two vertical lines is the same as
⟨Ψ|𝑄̂ Ψ⟩, i.e., ⟨Ψ|𝑄̂ |Ψ⟩ = ⟨Ψ|𝑄̂ Ψ⟩ since a hermitian operator can act forward or backwards on the state. If an
operator is not hermitian (does not correspond to a physical observable), one should assume that the operator acts
on the state after it (to the right of the operator) even if the notation ⟨Ψ|𝑄̂|Ψ⟩ is used.
27. Suppose a generic operator 𝑄̂ depends only on position and momentum operators. How is 𝑄̂ acting on a
generic state |Ψ⟩, i.e., 𝑄̂ |Ψ⟩, represented in the position representation when basis vectors are chosen to be
eigenstates of postion, |𝑥⟩?
𝜕
(I)
⟨𝑥|𝑄̂|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ ) |Ψ⟩
(II)
(III)
⟨𝑥|𝑄̂|Ψ⟩ =
(IV)
⟨𝑥|𝑄̂|Ψ⟩ =
(a)
(b)
(c)
(d)
𝜕𝑥
𝜕
)⟨𝑥|Ψ⟩
𝜕𝑥
𝜕
𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥)
𝜕
𝑥𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥)
⟨𝑥|𝑄̂|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ
(I) and (IV) only
(II) and (III) only
(I), (II), and (III) only
All of the above.
12
28. Consider the following conversation between two students:
 Student 1: The correct answer for question 27 is (b). It is just like question 26, except question 26 is
a special case of question 27.
𝑖𝑝′ 𝑥
𝑒 ℏ
𝜕
) similar to ⟨𝑥|𝑄̂|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥)?

Student 2: How is ⟨𝑥|𝑝̂ |𝑝′⟩ = 𝑝′ (

Student 1: Well, in question 26, ⟨𝑥|𝑝̂ |𝑝′⟩ is like ⟨𝑥|𝑄̂|Ψ⟩ because the operator 𝑝̂ corresponds to 𝑄̂ and
state |𝑝′⟩ corresponds to state |Ψ⟩. Both 𝑝̂ |𝑝′⟩ and 𝑄̂ |Ψ⟩ are operators acting on a state vector
without reference to a basis. When basis vectors are chosen to be eigenstates of position, |𝑥⟩, and we
take the scalar product of 𝑝̂ |𝑝′⟩ or 𝑄̂ |Ψ⟩ each with |𝑥⟩, we obtain the respective operators written in
position representation acting on the respective position space wavefunction. The operators and state
√2𝜋ℏ
𝑖𝑝′ 𝑥
vectors in each case are represented in position representation by
𝜕 𝑒 ℏ
−𝑖ℏ 𝜕𝑥 (
)
√2𝜋ℏ
and
𝑖𝑝′ 𝑥
𝜕
𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥). Also,
𝑒 ℏ
√2𝜋ℏ
= ⟨𝑥|𝑝′⟩, which is a momentum eigenfunction with eigenvalue 𝑝′ , is
a special type of position space wavefunction Ψ(𝑥) = ⟨𝑥|Ψ⟩.
Do you agree with Student 1’s explanation? Explain your reasoning.
29. Suppose a generic operator 𝑄̂ only depends on the position and momentum operators. How is 𝑄̂ acting on a
generic state |Ψ⟩, i.e., 𝑄̂ |Ψ⟩, representated in the momentum representation when basis vectors are chosen to
be eigenstates of momentum, |𝑝⟩?
(I)
(a)
(b)
(c)
(d)
𝜕
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑄̂ ( 𝑖ℏ 𝜕𝑝 , 𝑝) |Ψ⟩
(II)
𝜕
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑄̂ ( 𝑖ℏ 𝜕𝑝 , 𝑝) ⟨𝑝|Ψ⟩
(III)
𝜕
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑄̂ ( 𝑖ℏ 𝜕𝑝 , 𝑝) Φ(𝑝)
(IV)
𝜕
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑝𝑄̂ (𝑖ℏ 𝜕𝑝 , 𝑝) Φ(𝑝)
(I) and (II) only
(II) and (III) only
(II), (III), and (IV) only
All of the above
13
30. Consider the following conversation between Student A and Student B:
𝜕
 Student A: Is 𝑄̂ (𝑥, −𝑖ℏ ) the operator 𝑄̂ in the position representation, where 𝑄̂ is a function of position
𝜕𝑥
𝜕
and momentum operators written in position representation, i.e., 𝑥 and −𝑖ℏ 𝜕𝑥 , respectively?

Student B: Yes. Suppose 𝑄̂ is the kinetic energy operator. In position representation, 𝑄̂ will be
ℏ2
𝑝̂2
2𝑚
=
𝜕2
− 2𝑚 𝜕2 𝑥, if the particle is confined in one spatial dimension.

2
2
𝑝̂
𝑝
Student A: And if 𝑄̂ is the kinetic energy operator in momentum representation, 𝑄̂ will be 2𝑚 = 2𝑚, since
the momentum operator 𝑝̂ = 𝑝 in the momentum representation. In other words, in the momentum
representation, momentum operator 𝑝̂ acting on a state Φ(𝑝) simply amounts to multiplication of Φ(𝑝) by
𝑝 to yield 𝑝Φ(𝑝), i.e., 𝑝̂ Φ(𝑝) = 𝑝 Φ(𝑝).
Do you agree with Student A, Student B, both, or neither? Explain your reasoning.
 For a generic state |𝚿⟩, momentum space wavefunction 𝚽(𝒑) and position space wavefunction 𝚿(𝒙)
are a Fourier transform and inverse Fourier transform of each other, respectively.
Problem: Show that the momentum space wave function Φ(𝑝) is the Fourier transform of the position space
wave function  (x) .
The problem has been broken down into several multiple choice questions to guide your solution.
31. Which one of the following is the correct expression for Φ(𝑝)?
⟨𝑥|𝑝⟩
(a)
⟨𝑝|Ψ⟩
(b)
⟨𝑝|𝑥⟩
(c)
⟨Ψ|𝑝⟩
(d)
32. Using the completeness relation, insert a complete set of position eigenstates into the momentum space wave
function, ⟨𝑝|Ψ⟩. What do you obtain?
∞
(a)
∫−∞ 𝑑𝑥⟨𝑝|𝑝⟩⟨𝑝|Ψ⟩
∞
(c)
∫−∞ 𝑑𝑥⟨𝑝|𝑥⟩⟨𝑥|Ψ⟩
∞
∫−∞ 𝑑𝑝⟨𝑝|𝑥⟩⟨𝑥|Ψ⟩
(d)
∫−∞ 𝑑𝑝⟨𝑝|𝑝⟩⟨𝑝|Ψ⟩
(b)
∞
14
∞
∞
33. The expression after you insert the complete set 𝐼̂ = ∫−∞|𝑥⟩⟨𝑥|𝑑𝑥 should be ∫−∞ 𝑑𝑥⟨𝑝|𝑥⟩⟨𝑥|Ψ⟩. What is
⟨𝑝|𝑥⟩? You can go back to question 11 for help.
∞
34. Rewrite the expression ∫−∞ 𝑑𝑥⟨𝑝|𝑥⟩⟨𝑥|Ψ⟩, using your answer to the preceding question in place of ⟨𝑝|𝑥⟩.
35. Rewrite your answer to the preceding question replacing ⟨𝑥|Ψ⟩ with Ψ(𝑥), since Ψ(𝑥) = ⟨𝑥|Ψ⟩ is the most
common notation for the wave function in position representation.
1
∞
36. The Fourier transform of a general function 𝑓(𝑥) is 𝑔(𝑘) =
∫ 𝑒 −𝑖𝑘𝑥 𝑓(𝑥)𝑑𝑥. Compare this with your
√2𝜋 −∞
answer to the preceding question. The two expressions are analogous. (Hint: Using the de Broglie relation,
one can show that p  k , where p is the momentum and k is the wave number.)
15
Summary: Connecting Dirac notation with function space (position and momentum representation)
 Relationship between state vector |𝚿⟩ and the wavefunction
o Ψ(𝑥) = ⟨𝑥|Ψ⟩ is the representation of the state vector |Ψ⟩ in the position representation (when a
complete set of eigenstates of position |𝑥⟩ are chosen as basis vectors, i.e.,
∞
|Ψ⟩ = ∫−∞ Ψ(𝑥)|𝑥⟩𝑑𝑥) and is called the position space wavefunction.
o
Φ(𝑝) = ⟨𝑝|Ψ⟩ is the representation of the state vector |Ψ⟩ in the momentum representation (when
a complete set of eigenstates of momentum |𝑝⟩ are chosen as basis vectors, i.e.,
∞
|Ψ⟩ = ∫−∞ 𝛷(𝑝)|𝑝⟩𝑑𝑝) and is called the momentum space wavefunction.
o

The position space wavefunction Ψ(𝑥) and momentum space wavefunction Φ(𝑝) can be
represented as infinite dimensional column vectors in the position and momentum representation,
respectively.
o The momentum space wavefunction Φ(𝑝) = ⟨𝑝|Ψ⟩ and position space wavefunction Ψ(𝑥) =
⟨𝑥|Ψ⟩ are a Fourier transform and inverse Fourier transform of each other, respectively.
Position and momentum eigenstates
o The position eigenstates with eigenvalue 𝑥 ′ can be written as:
 |𝑥′⟩ in Dirac notation without reference to a basis.
 ⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ ) in position representation.
−𝑖𝑝𝑥′

o
⟨𝑝|𝑥′⟩ =
𝑒 ℏ
√2𝜋ℏ
in momentum representation.
The momentum eigenstates with eigenvalue 𝑝′ can be written as:
 |𝑝′⟩ in Dirac notation without reference to a basis.
𝑖𝑝′𝑥


⟨𝑥|𝑝′⟩ =
𝑒 ℏ
√2𝜋ℏ
in the position representation.
 ⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ ) in the momentum representation.
̂ |𝚿⟩ in position and momentum representations
𝑸
o For a generic state |Ψ⟩ and generic operator 𝑄̂ which depends only on 𝑥̂ and 𝑝̂ :
 𝑄̂ |Ψ⟩ is represented by ⟨𝑥|𝑄̂|Ψ⟩ in the position representation, which is by definition
𝜕
⟨𝑥|𝑄̂|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) Ψ(𝑥), where Ψ(𝑥) = ⟨𝑥|Ψ⟩.

𝑄̂ |Ψ⟩ is represented by ⟨𝑝|𝑄̂ |Ψ⟩ in the momentum representation, which is by definition
𝜕
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑄̂ (𝑖ℏ 𝜕𝑝 , 𝑝) Φ(𝑝), where Φ(𝑝) = ⟨𝑝|Ψ⟩.
o
Special cases of 𝑄̂ in position and momentum representations:
 𝑄̂ = 𝑥̂

𝜕

Position representation: ⟨𝑥|𝑥̂|Ψ⟩ = 𝑥̂(𝑥, −𝑖ℏ 𝜕𝑥)⟨𝑥|Ψ⟩ = 𝑥Ψ(𝑥)

Momentum representation: ⟨𝑝|𝑥̂|Ψ⟩ = 𝑥̂(𝑖ℏ 𝜕𝑝 , 𝑝)⟨𝑝|Ψ⟩ = 𝑖ℏ 𝜕𝑝 Φ(𝑝)
𝜕
𝜕
𝑄̂ = 𝑝̂
𝜕
𝜕

Position representation: ⟨𝑥|𝑝̂ |Ψ⟩ = 𝑝̂ (𝑥, −𝑖ℏ 𝜕𝑥)⟨𝑥|Ψ⟩ = −𝑖ℏ 𝜕𝑥 Ψ(𝑥)

Momentum representation: ⟨𝑝|𝑝̂ |Ψ⟩ = 𝑝̂ (𝑖ℏ 𝜕𝑝 , 𝑝)⟨𝑝|Ψ⟩ = 𝑝Φ(𝑝)
𝜕
16
Position eigenstates
(or position eigenfunctions)
Momentum eigenstates
(or momentum eigenfunctions)
Position representation
⟨𝑥|𝑥′⟩ = 𝛿(𝑥 − 𝑥 ′ )
⟨𝑝|𝑥′⟩ =
⟨𝑥|𝑝′⟩ =
Momentum operator 𝑝̂
−𝑖ℏ
𝜕
𝜕𝑥
Position Representation
𝑄̂ |𝑥′⟩
𝑄̂ |𝑝′⟩
𝑄̂ |Ψ⟩
⟨𝑥|𝑄̂ |𝑥 ′ ⟩ = 𝑄̂ (𝑥, −𝑖ℏ
𝜕
𝜕
) ⟨𝑥|𝑥 ′ ⟩ = 𝑄̂ (𝑥, −𝑖ℏ ) 𝛿(𝑥 − 𝑥 ′ )
𝜕𝑥
𝜕𝑥
𝜕
𝜕 𝑒 ℏ
= 𝑄̂ (𝑥, −𝑖ℏ ) ⟨𝑥|𝑝′ ⟩ = 𝑄̂ (𝑥, −𝑖ℏ )
𝜕𝑥
𝜕𝑥 √2𝜋ℏ
⟨𝑥|𝑄̂ |Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ
−𝑖𝑝𝑥 ′
ℏ
√2𝜋ℏ
Momentum Representation
𝜕
𝑖ℏ
𝜕𝑝
𝑝
Momentum Representation
𝑖𝑝′ 𝑥
⟨𝑥|𝑄̂ |𝑝′⟩
𝑒
√2𝜋ℏ
⟨𝑝|𝑝′⟩ = 𝛿(𝑝 − 𝑝′ )
𝑖𝑝′ 𝑥
𝑒 ℏ
Position Representation
𝑥
Position operator 𝑥̂
Momentum representation
𝜕
𝜕
) ⟨𝑥|Ψ⟩ = 𝑄̂ (𝑥, −𝑖ℏ ) Ψ(𝑥)
𝜕𝑥
𝜕𝑥
−𝑖𝑝𝑥 ′
𝜕
𝜕
𝑒 ℏ
⟨𝑝|𝑄̂ |𝑥 ′ ⟩ = 𝑄̂ (𝑖ℏ , 𝑝) ⟨𝑝|𝑥 ′ ⟩ = 𝑄̂ (𝑖ℏ , 𝑝)
𝜕𝑝
𝜕𝑝
√2𝜋ℏ
⟨𝑝|𝑄̂ |𝑝′ ⟩ = 𝑄̂ (𝑖ℏ
𝜕
𝜕
, 𝑝) ⟨𝑝|𝑝′⟩ = 𝑄̂ (𝑖ℏ , 𝑝) 𝛿(𝑝 − 𝑝′ )
𝜕𝑝
𝜕𝑝
⟨𝑝|𝑄̂ |Ψ⟩ = 𝑄̂ (𝑖ℏ
𝜕
𝜕
, 𝑝) ⟨𝑝|Ψ⟩ = 𝑄̂ (𝑖ℏ , 𝑝) Φ(𝑝)
𝜕𝑝
𝜕𝑝
17
Hamiltonian operator for a given quantum system and its eigenstates (energy eigenstates)
37. For a given quantum system, what is Ĥ acting on an energy eigenstate |𝜓𝑛 ⟩ (also represented by
|𝐸𝑛 ⟩ or |Ψ𝐸𝑛 ⟩) equal to?
̂ |𝜓𝑛 ⟩ = 𝐸|𝜓𝑛 ⟩, where E is the average of all possible energies of the system.
(a)
𝐻
̂ |𝜓𝑛 ⟩ = ∑𝑛 𝐸𝑛 |𝜓𝑛 ⟩, where En is the energy of the nth state.
(b)
𝐻
̂ |𝜓𝑛 ⟩ = 𝐸𝑛 |𝜓𝑛 ⟩, , where En is the energy of the nth state.
(c)
𝐻
(d)
None of the above. The Hamiltonian of a system must be known explicitly to determine the answer.
38. Consider the following conversation between Student A and Student B about the preceding question (the
symbols have the same meaning as in the preceding question):
̂ describes
 Student A: The answer should be choice (d) for the previous question. The Hamiltonian 𝐻
̂ explicitly.
the system, and we cannot determine what will happen without knowing 𝐻
 Student B: I agree. But this time we are given that |𝜓𝑛 ⟩ are the energy eigenstates. Energy
̂ . Any operator acting on one of its
eigenstates |𝜓𝑛 ⟩ are the eigenstates of the Hamiltonian 𝐻
eigenstates will give the same state back with the corresponding eigenvalue.
̂ of a system to calculate the energy
 Student A: Right. So we do need to know the Hamiltonian 𝐻
eigenstates |𝜓𝑛 ⟩ and eigenvalues 𝐸𝑛 .
̂ for a given quantum system to calculate the
 Student B: Yes. We would need the Hamiltonian 𝐻
energy eigenstates |𝜓𝑛 ⟩ and eigenvalues 𝐸𝑛 explicitly, but we are asked to select an expression for the
̂ . The expression in choice (c) is correct, even though we don’t know what
eigenvalue equation for 𝐻
̂.
the explicit |𝜓𝑛 ⟩’s and En’s are without knowing the Hamiltonian 𝐻
With whom do you agree? Explain.
18
39. Consider the following conversation between Student A and Student B:

̂ without choosing a basis (e.g., position or
Student A: The eigenvalue equation for a Hamiltonian 𝐻
̂ |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩. 𝐻
̂ |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩ can be written in position
momentum representation) is 𝐻
̂ (𝑥, −𝑖ℏ 𝜕 ) Ψ𝑛 (𝑥) = 𝐸𝑛 Ψ𝑛 (𝑥). We can see this by taking the inner product of
representation as 𝐻
𝜕𝑥
̂ |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩ with ⟨𝑥|, like this:
𝐻
̂
̂
⟨𝑥|𝐻 |Ψ𝑛 ⟩ = ⟨𝑥|𝐸𝑛 |Ψ𝑛 ⟩ . Using the fact that ⟨𝑥|𝑄|Ψ⟩
𝜕
̂ |Ψ𝑛 ⟩
= 𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥)Ψ(𝑥) for any generic operator 𝑄̂ and state |Ψ⟩, ⟨𝑥|𝐻
̂ (𝑥, −𝑖ℏ 𝜕 ) Ψ𝑛 (𝑥) = 𝐸𝑛 Ψ𝑛 (𝑥).
written as 𝐻
𝜕𝑥
𝜕
𝑄̂ (𝑥, −𝑖ℏ 𝜕𝑥) ⟨𝑥|Ψ⟩
⟨𝑥|𝐸𝑛 |Ψ𝑛 ⟩ can be

=
=
̂ as
Student B: I disagree with you. We can’t write the eigenvalue equation for the Hamiltonian 𝐻
̂
𝐻 |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩ because we don’t know the Hamiltonian explicitly. Also, we can’t write the
𝜕
̂ (𝑥, −𝑖ℏ 𝜕 ) if we don’t know the Hamiltonian
Hamiltonian in terms of 𝑥̂ and 𝑝̂ = −𝑖ℏ
as 𝐻
𝜕𝑥
𝜕𝑥
explicitly.
With whom do you agree? Explain your reasoning below.
19
Problem: Show that the expectation value of energy in a generic state |Ψ⟩ of the system is
̂ |Ψ⟩ = ∑𝑛|𝐶𝑛 |2 𝐸𝑛 , where 𝐶𝑛 (𝑛 = 1,2,3 … ∞) are the coefficients in the expansion |Ψ⟩ = ∑∞
⟨Ψ|𝐻
𝑛=1 𝐶𝑛 |Ψ𝑛 ⟩. En
th
is the eigenvalue for the n energy eigenstate |Ψ𝑛 ⟩.
The problem has been broken down into several multiple choice questions to guide your solution.
40. Student A uses the completeness relation and inserts a complete set of energy eigenstates |Ψ𝑛 ⟩, where 𝑛 =
̂ |Ψ⟩, so that it becomes ∑𝑛⟨Ψ|𝐻
̂ |𝜓𝑛 ⟩⟨𝜓𝑛 |Ψ⟩.
1,2,3 … ∞, into the expression for the expectation value ⟨Ψ|𝐻
̂ |𝜓𝑛 ⟩⟨𝜓𝑛 |Ψ⟩.
Choose all of the following statements that are correct about the new expression ∑𝑛⟨Ψ|𝐻
̂ |Ψ⟩, because ∑𝑛|𝜓𝑛 ⟩ ⟨𝜓𝑛 | is equal to the identity
(I)
It is the same as the expression ⟨Ψ|𝐻
operator.
̂ |Ψ⟩ and ∑𝑛⟨Ψ|𝐻
̂ |𝜓𝑛 ⟩⟨𝜓𝑛 |Ψ⟩ interchangeably based upon convenience
(II)
One can use ⟨Ψ|𝐻
because they are equal.
∑𝑛|𝜓𝑛 ⟩ ⟨𝜓𝑛 | is an operator, so it changes the expression when you insert it in the middle
(III)
̂ |Ψ⟩. Thus, the new expression is not the same as the old expression.
in ⟨Ψ|𝐻
(a)
(I) only
(b)
(II) only
(c)
(III) only
(d)
(I) and (II) only
41. Choose all of the following statements that are correct.
̂ |𝜓𝑛 ⟩⟨𝜓𝑛 |Ψ⟩ = ∑𝑛 𝐸𝑛 ⟨Ψ|𝜓𝑛 ⟩⟨𝜓𝑛 |Ψ⟩, because 𝐻
̂ |𝜓𝑛 ⟩ = 𝐸𝑛 |𝜓𝑛 ⟩ and 𝐸𝑛 is a
∑𝑛⟨Ψ|𝐻
(I)
number that can be pulled out of the inner product.
⟨Ψ|𝜓𝑛 ⟩ = ⟨𝜓𝑛 |Ψ⟩∗ .
(II)
(III)
𝐶𝑛 = ⟨𝜓𝑛 |Ψ⟩ (where |Ψ⟩ = ∑∞
𝑛=1 𝐶𝑛 |Ψ𝑛 ⟩).
(a)
(I) and (II) only
(b)
(I) and (III) only
(c)
(II) and (III) only
(d)
All of the above.
̂ |Ψ⟩ = ∑𝑛|⟨𝜓𝑛 |Ψ⟩|2 𝐸𝑛 = ∑𝑛|𝐶𝑛 |2 𝐸𝑛 .
42. Use your answers to the four preceding questions to show that ⟨Ψ|𝐻
20
Summary of the Hamiltonian operator for a given quantum system, its eigenstates, and expectation value of
energy:

̂ for a given system acting on an energy eigenstate state is
The result of the Hamiltonian operator 𝐻
̂ |Ψ𝑛 ⟩ = 𝐸𝑛 |Ψ𝑛 ⟩. In position representation, the eigenvalue
given by the eigenvalue equation 𝐻
̂ (𝑥, −𝑖ℏ 𝜕 ) Ψ𝑛 (𝑥) = 𝐸𝑛 Ψ𝑛 (𝑥).
equation for the Hamiltonian operator can be written as 𝐻

̂ |Ψ⟩ = ∑𝑛|𝐶𝑛 |2 𝐸𝑛 , where Cn = ⟨Ψ𝑛 |Ψ⟩
The expectation value of energy in a generic state |Ψ⟩ is ⟨Ψ|𝐻
th
is the coefficient in the expansion |Ψ⟩ = ∑∞
𝑛=1 𝐶𝑛 |Ψ𝑛 ⟩ and En is the eigenvalue of the n energy
eigenstate |Ψ𝑛 ⟩.
𝜕𝑥
Checkpoint
Consider the following statement about a particle in a one dimensional infinite square well:
a) If the state of the system at time 𝑡 = 0 is |Ψ⟩ =
1
[|Ψ1 ⟩ +
√2
|Ψ2 ⟩], in which |Ψ1 ⟩ and |Ψ2 ⟩ are the lowest
two energy eigenstates (ground state and first excited state with energy eigenvalues 𝐸1 and 𝐸2 ,
̂ |Ψ⟩ = 𝐸|Ψ⟩.
respectively), |Ψ⟩ satisfies the eigenvalue equation 𝐻
Explain why you agree or disagree with this statement.
̂ |Ψ⟩ for |Ψ⟩ =
b) Work out 𝐻
1
[|Ψ1 ⟩ +
√2
|Ψ2 ⟩] to check whether it satisfies the energy eigenvalue equation
̂ |Ψ⟩ = 𝐸|Ψ⟩.
𝐻
21
Answer to Checkpoint
a. Disagree. It is not an eigenstate of the Hamiltonian (not a stationary state).
̂ |Ψ⟩ = 𝐻
̂ ( 1 [|Ψ1 ⟩ + |Ψ2 ⟩]) =
b. 𝐻
2
√
1
[𝐸 |Ψ ⟩ +
√2 1 1
̂ ( 1 [|Ψ1 ⟩ + |Ψ2 ⟩]) ≠ 𝐸|Ψ⟩
𝐸2 |Ψ2 ⟩] ≠ 𝐸𝐻
2
√
22