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

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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 $[P,X^2]\neq 0$.

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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 $[e^{-\alpha a^+}, e^{-i\beta a^+a}]$.

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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 $[J,U] \neq 0$.

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+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

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