0
$\begingroup$

Consider the following defined commutation relations:

$$[\hat a,\hat a^{\dagger}]=1$$

$$[\hat b,\hat b^{\dagger}]=1$$

$$[\hat a,\hat b]=0$$

(where the usual algebra of commutators holds)

Let us now consider the operator:

$$\hat H=\hat a^{\dagger}\hat a+\hat b^{\dagger}\hat b+\hat a^{\dagger}\hat a(b^{\dagger}+b)+(a^{\dagger}+a)(b^{\dagger}+b)$$

I now want to evaluate the following using Mathematica: $$e^{\hat A}\hat He^{−\hat A}=[\hat A,\hat H]+[\hat A,[\hat A,\hat H]]+[\hat A,[\hat A,[\hat A,\hat H]]].....$$

where, $\hat A=(\hat a^{\dagger}-\hat a+\hat b^{\dagger}-\hat b)$

As can be seen, this is a really difficult task to perform by hand and I want to use Mathematica to do it. Since I am new at Mathematica, I don't even know how to begin.

Maybe I would have to somehow define the above relations as rules but I don't then know how to make Mathematica follow the commutator algebra rules.

I am not looking for a complete solution (though it's okay if it comes as an answer since I can always use it to match with my code:), just a guide on how to start this problem in Mathematica.

EDIT: From the link in the comment by evanb I realized that it's rather hard to do this in Mathematica. Has anybody used the SNEG package who can guide me on how to install or use it in Mathematica?

I am hoping that that should make my work easier. Since explicitly writing the code might take a very long time. I even have to change $\hat A$ in the above expression and try for many such $\hat A$'s.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Related: mathematica.stackexchange.com/q/95616/7936 $\endgroup$
    – evanb
    Commented Aug 5, 2021 at 12:28
  • $\begingroup$ Also: mathematica.stackexchange.com/q/63811/7936 $\endgroup$
    – evanb
    Commented Aug 5, 2021 at 12:33
  • $\begingroup$ The expression you're looking for is $e^\hat{A} \hat{H} e^{-\hat{A}}$ by the Baker-Hausdorff lemma. I assume that you already knew this, but if other non-physicists try to answer the question it might allow them to do an end run around calculating the entire series. (In particular, if you had a finite-dimensional representation of the operators, you could use MatrixExponential; but I suspect from the context that no such representation exists.) $\endgroup$
    – Michael Seifert
    Commented Aug 5, 2021 at 15:09
  • $\begingroup$ @Michael Seifert Yes. I have directly written the expansion of the above. I will edit and add the expression above. Also, I think the name of the above expansion is Hadamard Lemma. Baker-Hausdorff lemma is for the expansion of $e^{\hat A}e^{\hat B}$ I think. $\endgroup$
    – Lost
    Commented Aug 5, 2021 at 15:12
  • 3
    $\begingroup$ It's called the "Baker-Hausdorff lemma" in Sakurai's Modern Quantum Mechanics (not to be confused with the Baker-Campbell-Hausdorff formula, which gives us an expansion for $e^\hat{A} e^\hat{B}$.) But it's entirely possible that there are other names for it. $\endgroup$
    – Michael Seifert
    Commented Aug 5, 2021 at 15:16

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.