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Theme 2 — Hilbert Spaces and Dirac Notation

Properties of commutators

Keywords: commutator · Leibniz rule · Jacobi identity · anti-Hermitian operator · trace · canonical commutation relation · Pauli matrices

Exercise 1 : Properties of commutators

For two operators A^\hat A and B^\hat B on a finite-dimensional Hilbert space, the commutator is defined by [A^,B^]=A^B^−B^A^[\hat A,\hat B]=\hat A\hat B-\hat B\hat A.

Question 1
Show that the commutator is antisymmetric and bilinear, then establish the Leibniz rule [A^,B^C^]=B^[A^,C^]+[A^,B^]C^.[\hat A,\hat B\hat C]=\hat B[\hat A,\hat C]+[\hat A,\hat B]\hat C .

Solution
Antisymmetry, [B^,A^]=−[A^,B^][\hat B,\hat A]=-[\hat A,\hat B], and bilinearity follow directly from the definition. For the Leibniz rule, expand the right-hand side:

B^(A^C^−C^A^)+(A^B^−B^A^)C^=B^A^C^−B^C^A^+A^B^C^−B^A^C^=A^B^C^−B^C^A^,\hat B(\hat A\hat C-\hat C\hat A)+(\hat A\hat B-\hat B\hat A)\hat C=\hat B\hat A\hat C-\hat B\hat C\hat A+\hat A\hat B\hat C-\hat B\hat A\hat C=\hat A\hat B\hat C-\hat B\hat C\hat A,

which is indeed [A^,B^C^][\hat A,\hat B\hat C]. The commutator with A^\hat A acts on a product as a derivative acts on a product of functions, except that the order of the factors must be respected.

Question 2
Establish the Jacobi identity: [A^,[B^,C^]]+[B^,[C^,A^]]+[C^,[A^,B^]]=0[\hat A,[\hat B,\hat C]]+[\hat B,[\hat C,\hat A]]+[\hat C,[\hat A,\hat B]]=0.

Solution
Expand the first term: [A^,[B^,C^]]=A^B^C^−A^C^B^−B^C^A^+C^B^A^[\hat A,[\hat B,\hat C]]=\hat A\hat B\hat C-\hat A\hat C\hat B-\hat B\hat C\hat A+\hat C\hat B\hat A. The other two terms are obtained by the cyclic permutation A^→B^→C^→A^\hat A\to\hat B\to\hat C\to\hat A:

[B^,[C^,A^]]=B^C^A^−B^A^C^−C^A^B^+A^C^B^,[C^,[A^,B^]]=C^A^B^−C^B^A^−A^B^C^+B^A^C^.[\hat B,[\hat C,\hat A]]=\hat B\hat C\hat A-\hat B\hat A\hat C-\hat C\hat A\hat B+\hat A\hat C\hat B, \qquad [\hat C,[\hat A,\hat B]]=\hat C\hat A\hat B-\hat C\hat B\hat A-\hat A\hat B\hat C+\hat B\hat A\hat C .

Adding the three lines, each of the six products appears once with a ++ sign and once with a −- sign: the sum vanishes.

Question 3
Show that if A^\hat A and B^\hat B are self-adjoint, their commutator is anti-self-adjoint, [A^,B^]†=−[A^,B^][\hat A,\hat B]^\dagger=-[\hat A,\hat B], and that i[A^,B^]i[\hat A,\hat B] is self-adjoint. Deduce that the expectation value ⟨ψ∣[A^,B^]∣ψ⟩\bra\psi[\hat A,\hat B]\ket\psi is purely imaginary.

Solution
Taking the adjoint of a product reverses the order of the factors: (A^B^)†=B^†A^†=B^A^(\hat A\hat B)^\dagger=\hat B^\dagger\hat A^\dagger=\hat B\hat A. Thus [A^,B^]†=B^A^−A^B^=−[A^,B^][\hat A,\hat B]^\dagger=\hat B\hat A-\hat A\hat B=-[\hat A,\hat B]. Consequently, (i[A^,B^])†=−i⋅(−[A^,B^])=i[A^,B^](i[\hat A,\hat B])^\dagger=-i\cdot(-[\hat A,\hat B])=i[\hat A,\hat B]: this operator is self-adjoint, and its expectation value is real. The expectation value of [A^,B^][\hat A,\hat B] is therefore −i-i times a real number, that is, purely imaginary. This is why commutation relations between observables involve the factor ii, as in [X^,P^]=iℏ1[\hat X,\hat P]=i\hbar\mathbf{1} or [S^x,S^y]=iℏS^z[\hat S_x,\hat S_y]=i\hbar\hat S_z, and also why the Robertson inequality involves the modulus of ⟨[A^,B^]⟩\langle[\hat A,\hat B]\rangle.

Question 4
Calculate [σx,σy][\sigma_x,\sigma_y], with

σx=(0110),σy=(0−ii0),σz=(100−1).\begin{aligned} \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad\sigma_y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix},\qquad\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \end{aligned}

Verify the property from the previous question.

Solution
Calculate the two products:

σxσy=(0110)(0−ii0)=(i00−i)=iσz,σyσx=(−i00i)=−iσz.\begin{aligned} \sigma_x\sigma_y=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\ i&0\end{pmatrix}=\begin{pmatrix}i&0\\0&-i\end{pmatrix}=i\sigma_z, \qquad \sigma_y\sigma_x=\begin{pmatrix}-i&0\\0&i\end{pmatrix}=-i\sigma_z . \end{aligned}

Thus [σx,σy]=2iσz[\sigma_x,\sigma_y]=2i\sigma_z. This commutator is indeed anti-self-adjoint, since (2iσz)†=−2iσz(2i\sigma_z)^\dagger=-2i\sigma_z, and i[σx,σy]=−2σzi[\sigma_x,\sigma_y]=-2\sigma_z is self-adjoint.

Question 5
Show that tr⁡[A^,B^]=0\tr[\hat A,\hat B]=0. Deduce that there are no finite-dimensional matrices XX and PP satisfying [X,P]=iℏ1[X,P]=i\hbar\mathbf{1}.

Solution
The trace of a product does not depend on the order of its factors: tr⁡(A^B^)=tr⁡(B^A^)\tr(\hat A\hat B)=\tr(\hat B\hat A), hence tr⁡[A^,B^]=0\tr[\hat A,\hat B]=0. If [X,P]=iℏ1[X,P]=i\hbar\mathbf{1} held for n×nn\times n matrices, the trace of the left-hand side would vanish, while that of the right-hand side would be iℏn≠0i\hbar n\neq0. This is impossible: the canonical commutation relation between position and momentum can only be realised in an infinite-dimensional Hilbert space, where at least one of the two operators is necessarily unbounded.