Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say there is $\hat{J} = \exp[-i \hat{p} l/ \hbar]$ and $\hat{U}= \exp[-i\hat{H}t/ \hbar]$, where $\hat{H}$ is time-independent.

Can we say anything about $[\hat{J},\hat{U}]$? Is it zero? How do we show this?

For example if $\hat{H} = \hat{p}^2 /2m + m\omega^2 \hat{x}^2/2$.

share|cite|improve this question
Hint for things like this: you can drop any function of an operator like $\exp$ when testing commutativity. $U, J$ commute iff $p, H$ do. But see Joshphysics's answer and the little lemma that he proves: this one involves commutation of things with $U$ and is often stated in that form. – WetSavannaAnimal aka Rod Vance Oct 2 '13 at 11:57

It depends on the Hamiltonian.

In general in quantum mechanics, if $V$ is a unitary operator representing some symmetry, then we say that $H$ is invariant under that symmetry provided the Hamiltonian is invariant under conjugation by $V$; \begin{align} V^{-1} H V = H. \end{align} Notice that this condition can also be written as $[H,V]=0$. Now, if a Hamiltonian is invariant under such a symmetry, then we can multiply both sides by $-it/\hbar$, take the operator exponential of both sides, and use the identity $e^{A^{-1}BA} = A^{-1}e^BA$ to obtain \begin{align} V^{-1}UV = U \end{align} which can be written as \begin{align} [U,V]=0. \end{align} On the other hand, suppose that $[U,V]=0$, then expand the commutator in powers of $t$. This gives \begin{align} [I -(it/\hbar)H +\cdots, V]=0 \end{align} which, after equating all coefficients of powers of $t$ on the left to zero implies $[H,V]=0$. So we have shown that

The hamiltonian is invariant under a symmetry $V$, if and only if the time evolution operator $U$ commutes with $V$.

In the special case of spatial translations $T$ (which you have rather non-standardly labeled as $J$), the property $[U,T]=0$ holds if and only if $H$ is translation-invariant.

In the case of $H=P^2/2m+m\omega^2X^2$, the Hamiltonian is not translation-invariant because \begin{align} T^{-1}HT = H + \frac{it}{\hbar}[P,H] + O(t^2) \end{align} and $[P,H]\neq 0$ for this Hamiltonian since $[P,P^2]=0$ but $[P,X^2]\neq 0$.

share|cite|improve this answer
Thank you for the great response, but your identity $e^{A^{-1}BA} = A^{-1}e^BA$ confused me I don't think I've ever run across it. Would you know where I could find a proof, or citation? Also in this case, the translation operator is an observable yes? What would the expectation value mean physically? – walczyk Oct 2 '13 at 16:48
Proving the identity: you can do it yourself. Just write out $e^{A^{-1}BA}$ in its power series form, realize that due to $A^{-1}A = 1$ what you're left is $A^{-1}$ and $A$ sandwiching the power series of $e^B$, so you have it. – nervxxx Oct 2 '13 at 17:05
@walczyk Your next question: the translation operator is not an observable. It is not hermitian, thus failing the axiom of QM that all observables are represented by hermitian operators. – nervxxx Oct 2 '13 at 17:07
@nervxxx Ah, I thought it was unitary, and therefore hermitian. I thought that if p was hermitian then so would $e^{\alpha p}$ What is a proper test for hermiticity? – walczyk Oct 2 '13 at 17:27
@walczyk you forget that under complex conjugating, $\alpha \to \alpha^*$. That's what makes it non-hermitian. – nervxxx Oct 2 '13 at 18:33

Note that in the very particular case of the quantum harmonic oscillator, it is interesting to use different representations. For instance:

$$P \sim i(a-a^+), H \sim a^+a \tag{1}$$

You have formulae which allow you to disentangle $a$ and $a^+$ (below $\lambda$ is real):

$$e^{\lambda(a^+-a)}=e^{\lambda a^+} e^{-\lambda a} e^{- \frac{\lambda^2}{2} }\tag{2}$$

This reduces the problem of finding a commutator $[e^{-i\alpha P}, e^{-i\beta H}]$, to finding commutators $[e^{\alpha a}, e^{-i\beta a^+a}]$ and $[e^{-\alpha a^+}, e^{-i\beta a^+a}]$.

share|cite|improve this answer
How would you find those commutators? Other than expanding them I mean. Thank you for pointing out representing them as the ladder operators. I notice our operator becomes $e^{-|\lambda|^2}e^{(\lambda a^\dagger)^\dagger}e^{-(\lambda a)^\dagger}e^{\beta a^\dagger a}e^{\lambda a^\dagger}e^{-\lambda a}$ Does that seem right to you? What would be the next step then? – walczyk Oct 2 '13 at 17:16
Good question...I will think to this, unless you ask a new PSE question, which is maybe the quicker way to have an answer of somebody with better skills than me. I don't understand you last expression, which operator is this ? – Trimok Oct 2 '13 at 17:24
Not also that you may use the notation $N = a^+a$, with $[N,a]= -a$, and $[N,a^+]= a^+$ – Trimok Oct 2 '13 at 17:30
Ah, my mistake! I wrote out $T^\dagger U T$ where T is our translation operator, and U is our time translation operator. I switched the two. It's then like you said $e^{|\lambda|^2}e^{(\beta a^\dagger a)^\dagger}e^{\lambda a^\dagger}e^{-\lambda a}e^{(\beta a^\dagger a)}$. – walczyk Oct 2 '13 at 17:32
I don't know what your expression represents, but it is not a commutator. – Trimok Oct 2 '13 at 17:37

This is a good formula to remember, or at least, to think of, when you're dealing with the exponential of operators:

Baker–Campbell–Hausdorff formula

In particular, for your case, the braiding identity is useful. We see that $[J,U] \neq 0$.

share|cite|improve this answer
+1 Just a nitpick from a Lie theory enthusiast: as generally proven, CBH applies to finite dimensional operators. So you have to make appropriate boundedness assumptions to make CBH work in quantum operator contexts. Also, (this is getting pedantic, but I like history) its the CBH theorem: none of Campbell, Baker or Hausdorff came up with a formula -they simply proved the series involves only Lie brackets that converges for some neighbourhood of the identity - this is all that is needed mostly for Lie theory. They of course came up with quite a few of the first co-efficients. But the actual .. – WetSavannaAnimal aka Rod Vance Oct 2 '13 at 11:48
..formula, as given in Rossmann Ch 1. (and is almost unreadable it's so complicated - I once had to code it into software and had a hard time even understanding what I was doing with the formula), is owing to (the still living) Eugene Dynkin. The "Braiding Identity" is not actually a CBH formula, although it does often seem to be presented alongside in Lie theory texts. This one is universally convergent, so much more straightforward. BTW I never did learn why it gets called the enchanting name of "braiding identity" - do you know? – WetSavannaAnimal aka Rod Vance Oct 2 '13 at 11:50
Thanks for the link. It was in the back of my mind, but it wasn't really simplifying anything for me. – walczyk Oct 2 '13 at 16:57
@WetSavannaAnimalakaRodVance thanks for the history lesson! no idea about the name of the identity. Possibly because when you try to move $e^X$ past $e^Y$ you get a tangle of commutators of $Y$ and $X$ ('braid') in $\cdots$ of $e^{Y+\cdots} e^X$... – nervxxx Oct 2 '13 at 17:12
I rewrote $T^\dagger U T$ using the suggestion, but I was wondering if you could see the next step. $T\dagger U T = e^{|\lambda|^2 /2} e^{(\beta a^\dagger a)^\dagger}e^{\lambda a^\dagger}e^{-\lambda a} e^{\beta a^\dagger a} = e^{|\lambda|^2 /2} e^{\beta^\dagger a^\dagger a}e^{\lambda a^\dagger}e^{-\lambda a} e^{\beta a^\dagger a}$ $ = e^{|\lambda|^2 /2}e^{\lambda a^\dagger e^{\beta^\dagger}}e^{\beta^\dagger a^\dagger a} e^{\beta a^\dagger a + \beta a (e^{-\lambda} - 1)}e^{-\lambda a}$ Any suggestions? – walczyk Oct 2 '13 at 17:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.