Quantum commutation relation $[e^{-x^{2}},e^{\alpha i p}] = ?$

Question

I have been trying for find a closed form solution, or at least something neat for the commutation relation $$[e^{-x^{2}},e^{\alpha i p}] = ?$$ (where $[x,p] = i\mathbb{I}$) but have had little luck. I have tried using BHC theorem but this does not get me very far. I think that there must be some simple relation that I am over looking.

Does this answer your question? Commutators involving functions — ZeroTheHero, Jun 15 '22 at 12:06
Oops I made a mistake. Let me fix the commutator. As it is prior to the changes indeed that post that you shared would suffice. — Hldngpk, Jun 15 '22 at 12:25
@Photon there, this is the commutator I was interested in. I accidentaly wrote down something else. — Hldngpk, Jun 15 '22 at 12:27
Here I have used BCH theorem and them I must compute interations of nested commutators involving the operator $p$. Using the stackexchange post that you shared I end up with a messy series of operators. I was hoping that there would be a nice closed form for the commutator posted above that does not need BCH theorem. — Hldngpk, Jun 15 '22 at 12:29

Cosmas Zachos · Accepted Answer · 2022-06-16T14:29:24.923

7

You have set $\hbar=1$, so $p= -i\partial_x$ in the coordinate representation, so one of your operators is a bland Lagrange shift operator, and hence $$ [e^{\alpha i p}, f(x) ] = (f(x+\alpha)-f(x)) e^{\alpha i p} ~~~~\leadsto \\ [e^{-x^2}, e^{\alpha i p}]= - (e^{-(x+\alpha)^2}-e^{x^2}) e^{\alpha i p} . $$

(With a tip of the hat to @thedude 's comment! The linked WP article reminds you that $e^{i\alpha p} f(x) e^{-i\alpha p}= f(x+\alpha) $, in operator calculus language; when it acts on a constant, it reduces to just $e^{i\alpha p} f(x)= f(x+\alpha) $.)

Make sure to confirm for small α.

edited Jun 16 '22 at 14:29

answered Jun 15 '22 at 13:19

Cosmas Zachos

62,595

This is phenomenal !!! Thank you very much! – Hldngpk Jun 15 '22 at 13:22
Actually $[e^{\alpha ip},f(x)]g(x)=e^{\alpha ip}f(x)g(x)-f(x)e^{\alpha ip}g(x)$, so this answer is in the right direction but not correct – thedude Jun 15 '22 at 13:22
Do you know where I might find a proof of these wonderful result. i.e. $[e^{\alpha i p},f(x)] = f(x+\alpha)$ where $x$ and $p$ are the position and momentum operators respectively. – Hldngpk Jun 15 '22 at 13:23
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After the edit, this is now correct – thedude Jun 15 '22 at 13:34
Fantastic! I appreciate all of your comments and help. Special thanks to @CosmasZachos – Hldngpk Jun 15 '22 at 13:35
The answer is now correct. Your last comment is correct only if you look at $e^{i \alpha p}f(x)$ as an operator. But if the operator is $e^{i \alpha p}$ it transforms the function $f(x)$ in the function $f(x+a)$. In any case, the final results can be obtained independently of this alternative views. – GiorgioP-DoomsdayClockIsAt-90 Jun 15 '22 at 13:49
@GiorgioP yes, of course. I edited the question to accommodate action on a translationally invariant state such as the vacuum. The combinatoric derivation is the standard matrix exponential derivation, but I thought Lagrange's argument is self-evident for those who have not encountered it. – Cosmas Zachos Jun 15 '22 at 13:53

Quantum commutation relation $[e^{-x^{2}},e^{\alpha i p}] = ?$

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