Commutation Relations

1. In Depth Insights

The commutator between two linear operators

A

and

B

is defined as:

[A,B] \equiv AB - BA

[A, B] = 0

, then

AB = BA

, meaning that operator

A

commutes with operator

B

. Similarly, the anti commutator of two operators is defined as:

\{A,B\} \equiv AB + BA

When

{A, B} = 0

, the operator

A

is said to anti commute with

B

. Many vital relations in quantum mechanics can be deduced directly using commutators and anti commutators.

Perhaps the most practically useful property of the commutator is its direct link to simultaneous diagonalization: two Hermitian operators

A

and

B

commute if and only if they can be simultaneously diagonalized by a shared orthonormal basis. That is, there exists a complete set of orthonormal eigenvectors

|vᵢ⟩

such that:

A := \sum_i a_i \lvert v_i\rangle\langle v_i\rvert

and

B := \sum_i b_i \lvert v_i\rangle\langle v_i\rvert

Simultaneous Diagonalization Theorem: Let

A

and

B

be Hermitian operators. Then

[A, B] = 0

if and only if there exists an orthonormal basis in which both

A

and

B

are diagonal. In this case, we say that

A

and

B

are simultaneously diagonalizable.

This foundational result connects the commutator, which is typically straightforward to evaluate directly, with the capacity for simultaneous diagonalization, a property that is generally far harder to determine a priori.

For example, consider the Pauli matrices

X

and

Y

[X,Y] = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} - \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} = 2i \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} = 2iZ

Therefore,

X

and

Y

do not commute.

Proof: One can easily prove that if

A

and

B

are diagonal in the same orthonormal basis, then

[A, B] = 0

. To prove the converse, let

|a, j⟩

be an orthonormal basis for the eigenspace

V_a

corresponding to the eigenvalue

a

of operator

A

, where index

j

resolves any degeneracy. Notice that:

AB∣a,j⟩=BA∣a,j⟩=aB∣a,j⟩,

Thus, the state

B |a, j⟩

is also an element of the eigenspace

V_a.

Let

P_a

denote the projection operator onto the subspace

V_a,

and define the operator

B_a ≡ P_a B P_a.

It is evident that the restriction of

B_a

to the subspace

V_a

is Hermitian, thereby admitting a spectral decomposition in terms of an orthonormal set of eigenvectors spanning

V_a

Let us denote these shared eigenvectors by

|a, b, k⟩

, where indices

a

and

b

label the eigenvalues of

A

and

B_a

, respectively, and

k

accounts for any residual degeneracy of

B_a

. Because

B |a, b, k⟩

resides in

V_a

, we have

B∣a,b,k⟩=P_{a}​B∣a,b,k⟩

. Furthermore, since

Pa​∣a,b,k⟩=∣a,b,k⟩

, we obtain:

B∣a,b,k⟩= Pa B Pa ∣a,b,k⟩=b∣a,b,k⟩.

Consequently,

|a, b, k⟩

is an eigenvector of

B

with eigenvalue

b

. It follows that the states

|a, b, k⟩

form a complete orthonormal basis of eigenvectors for both

A

and

B

across the entire vector space, establishing that

A

and

B

are simultaneously diagonalizable.

2. Mathematical Formalism

Consider the time independent Schrödinger equation for a simple harmonic oscillator:

\frac{1}{2m}\left[\hat{p}^{\,2} + (m\omega x)^2\right]\psi = E\psi

where

\hat{p} \equiv -i\hbar \frac{d}{dx}

represents the quantum momentum operator. The underlying strategy is to factor the Hamiltonian:

\hat{H}=\frac{1}{2m}\left[\hat{p}^{\,2}+(m\omega x)^2\right]

If these variables were ordinary numbers, the factorization would be elementary:

u^2 + v^2 = (iu + v)(-iu + v)

In quantum mechanics, however, this algebraic simplification is invalid because

x̂

and

p̂

are operators that generally do not commute; that is,

x̂p̂

is not equal to

p̂x̂

. Nonetheless, this inspires us to introduce the following ladder operators:

\hat{a}_{\pm} \equiv \frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\hat{p} + m\omega x\right)

What algebraic structure do these operators

â_±

possess? Let us calculate their product:

\hat{a}_{-}\hat{a}_{+} = \frac{1}{2\hbar m\omega} \left(i\hat{p}+m\omega x\right) \left(-i\hat{p}+m\omega x\right) = \frac{1}{2\hbar m\omega} \left[ \hat{p}^{\,2}+(m\omega x)^2-im\omega\left(x\hat{p}-\hat{p}x\right) \right].

As anticipated, an additional cross term emerges involving

(x̂p̂ - p̂x̂)

. This expression is precisely the commutator of

x̂

and

p̂

, which quantifies the degree to which the two operators fail to commute. In commutator notation, the product reads:

\hat{a}_{-}\hat{a}_{+} = \frac{1}{2\hbar m\omega} \left[ \hat{p}^{\,2}+(m\omega x)^2 \right] - \frac{i}{2\hbar}[x,\hat{p}]

To evaluate the commutator

[x, p̂]

, we must exercise care: working with abstract differential operators directly can lead to algebraic slips. The proper rigorous procedure is to let the commutator act upon an arbitrary test function

f(x)

[x,\hat{p}]\,f(x) = \left[ x(-i\hbar)\frac{d}{dx}(f) - (-i\hbar)\frac{d}{dx}(xf) \right] = -i\hbar \left( x\frac{df}{dx} - x\frac{df}{dx} - f \right) = i\hbar f(x).

Dropping the test function because it holds for any arbitrary smooth function, we arrive at the canonical commutation relation:

[x,\hat{p}] = i\hbar

This canonical commutation relation constitutes one of the most fundamental identities in quantum mechanics.

3. References

Nielsen, M. A., & Chuang, I. L. (2000). Quantum Computation and Quantum Information (Vol. 1). Cambridge University Press.

1. In Depth Insights

The commutator between two linear operators

A

and

B

is defined as:

[A,B] \equiv AB - BA

[A, B] = 0

, then

AB = BA

, meaning that operator

A

commutes with operator

B

. Similarly, the anti commutator of two operators is defined as:

\{A,B\} \equiv AB + BA

When

{A, B} = 0

, the operator

A

is said to anti commute with

B

. Many vital relations in quantum mechanics can be deduced directly using commutators and anti commutators.

Perhaps the most practically useful property of the commutator is its direct link to simultaneous diagonalization: two Hermitian operators

A

and

B

commute if and only if they can be simultaneously diagonalized by a shared orthonormal basis. That is, there exists a complete set of orthonormal eigenvectors

|vᵢ⟩

such that:

A := \sum_i a_i \lvert v_i\rangle\langle v_i\rvert

and

B := \sum_i b_i \lvert v_i\rangle\langle v_i\rvert

Simultaneous Diagonalization Theorem: Let

A

and

B

be Hermitian operators. Then

[A, B] = 0

if and only if there exists an orthonormal basis in which both

A

and

B

are diagonal. In this case, we say that

A

and

B

are simultaneously diagonalizable.

For example, consider the Pauli matrices

X

and

Y

[X,Y] = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} - \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} = 2i \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} = 2iZ

Therefore,

X

and

Y

do not commute.

Proof: One can easily prove that if

A

and

B

are diagonal in the same orthonormal basis, then

[A, B] = 0

. To prove the converse, let

|a, j⟩

be an orthonormal basis for the eigenspace

V_a

corresponding to the eigenvalue

a

of operator

A

, where index

j

resolves any degeneracy. Notice that:

AB∣a,j⟩=BA∣a,j⟩=aB∣a,j⟩,

Thus, the state

B |a, j⟩

is also an element of the eigenspace

V_a.

Let

P_a

denote the projection operator onto the subspace

V_a,

and define the operator

B_a ≡ P_a B P_a.

It is evident that the restriction of

B_a

to the subspace

V_a

is Hermitian, thereby admitting a spectral decomposition in terms of an orthonormal set of eigenvectors spanning

V_a

Let us denote these shared eigenvectors by

|a, b, k⟩

, where indices

a

and

b

label the eigenvalues of

A

and

B_a

, respectively, and

k

accounts for any residual degeneracy of

B_a

. Because

B |a, b, k⟩

resides in

V_a

, we have

B∣a,b,k⟩=P_{a}​B∣a,b,k⟩

. Furthermore, since

Pa​∣a,b,k⟩=∣a,b,k⟩

, we obtain:

B∣a,b,k⟩= Pa B Pa ∣a,b,k⟩=b∣a,b,k⟩.

Consequently,

|a, b, k⟩

is an eigenvector of

B

with eigenvalue

b

. It follows that the states

|a, b, k⟩

form a complete orthonormal basis of eigenvectors for both

A

and

B

across the entire vector space, establishing that

A

and

B

are simultaneously diagonalizable.

2. Mathematical Formalism

Consider the time independent Schrödinger equation for a simple harmonic oscillator:

\frac{1}{2m}\left[\hat{p}^{\,2} + (m\omega x)^2\right]\psi = E\psi

where

\hat{p} \equiv -i\hbar \frac{d}{dx}

represents the quantum momentum operator. The underlying strategy is to factor the Hamiltonian:

\hat{H}=\frac{1}{2m}\left[\hat{p}^{\,2}+(m\omega x)^2\right]

If these variables were ordinary numbers, the factorization would be elementary:

u^2 + v^2 = (iu + v)(-iu + v)

In quantum mechanics, however, this algebraic simplification is invalid because

x̂

and

p̂

are operators that generally do not commute; that is,

x̂p̂

is not equal to

p̂x̂

. Nonetheless, this inspires us to introduce the following ladder operators:

\hat{a}_{\pm} \equiv \frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\hat{p} + m\omega x\right)

What algebraic structure do these operators

â_±

possess? Let us calculate their product:

\hat{a}_{-}\hat{a}_{+} = \frac{1}{2\hbar m\omega} \left(i\hat{p}+m\omega x\right) \left(-i\hat{p}+m\omega x\right) = \frac{1}{2\hbar m\omega} \left[ \hat{p}^{\,2}+(m\omega x)^2-im\omega\left(x\hat{p}-\hat{p}x\right) \right].

As anticipated, an additional cross term emerges involving

(x̂p̂ - p̂x̂)

. This expression is precisely the commutator of

x̂

and

p̂

, which quantifies the degree to which the two operators fail to commute. In commutator notation, the product reads:

\hat{a}_{-}\hat{a}_{+} = \frac{1}{2\hbar m\omega} \left[ \hat{p}^{\,2}+(m\omega x)^2 \right] - \frac{i}{2\hbar}[x,\hat{p}]

To evaluate the commutator

[x, p̂]

, we must exercise care: working with abstract differential operators directly can lead to algebraic slips. The proper rigorous procedure is to let the commutator act upon an arbitrary test function

f(x)

[x,\hat{p}]\,f(x) = \left[ x(-i\hbar)\frac{d}{dx}(f) - (-i\hbar)\frac{d}{dx}(xf) \right] = -i\hbar \left( x\frac{df}{dx} - x\frac{df}{dx} - f \right) = i\hbar f(x).

Dropping the test function because it holds for any arbitrary smooth function, we arrive at the canonical commutation relation:

[x,\hat{p}] = i\hbar

This canonical commutation relation constitutes one of the most fundamental identities in quantum mechanics.

3. References

Nielsen, M. A., & Chuang, I. L. (2000). Quantum Computation and Quantum Information (Vol. 1). Cambridge University Press.

Commutation Relations

Article headings

1. In Depth Insights

2. Mathematical Formalism

3. References

Related concepts

ماتریس چگالی

Quantum Computing

Quantum Gates

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Commutation Relations

Article headings

1. In Depth Insights

2. Mathematical Formalism

3. References

Related concepts

ماتریس چگالی

Quantum Computing

Quantum Gates