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I need to compute this:

$$ [ e^{-\sigma K_+}, K_{-} ] $$ and $K_{\pm} $ has the following commutator relations

$$ [K_0, K_{\pm} ] = \pm K_{\pm}$$

$$ [K_{-}, K_{+} ] = 2K_0 .$$

I tried to expand the exponential operator and compute the interated commutator

$$ [ e^{-\sigma K_+}, K_{-} ] = \sum_0^\infty \frac{(-\sigma)^n}{n!} [K_{+}^n, K_{-} ] = 2 \sigma K_0 - 2 \sigma^2 K_{-} $$

This result gives zero, but if we expand the sum, we see that only surviving the two first terms.

My question is when can you use this formula? :

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and this is different from this:

enter image description here

I take BCH relations from wiki. https://en.wikipedia.org/wiki/Unitary_transformation_(quantum_mechanics)

I really appreciate your help, my final goal is to calculate this:

enter image description here

Qmechanic
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PCat27
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2 Answers2

1

  1. Recall first that the adjoint representation is given as $$({\rm ad} X)Y~:=~[X,Y].\tag{1}$$

  2. We can then write nested commutators as $$({\rm ad} X)^nY.\tag{2}$$

  3. Hadamard's formula can then be stated as $$ e^XYe^{-X}~=~e^{{\rm ad} X}Y. \tag{3}$$

  4. OP and Wikipedia instead adapt the notation $$ [(X)^n, Y]\tag{4}$$ for the nested commutators (2), which is confusing, since (2) is not the same as $$X^n Y -Y X^n ,\tag{5}$$ if $n>1$. Presumably the core of OP's questions is caused by the use of the confusing notation (4).

  5. For OP's squeezed state formula at the end, my Phys.SE answer here might be helpful.

Qmechanic
  • 201,751
1

The essence of the BCH identity and the Hadamard Lemma ((3) of the answer of @Qmechanic, above) you are quoting is that commutators make most sense for elements of the Lie algebra, which your mixed commutator of a group and algebra element is evidently not.

So your first task it to clean it up into a product of two such: $$ [ e^{-\sigma K_+}, K_{-} ]= ( e^{-\sigma K_+} K_{-} e^{\sigma K_+}- K_{-} ) ~~~e^{-\sigma K_+}. $$ Now, as per your Hadamard Lemma of WP, this is a product of the algebra element in the parenthesis, times the group element (the exponential).

Section III of this paper reminds you of the basic technique involving faithful representations of Lie algebras (not groups!), such as the Pauli-matrix doublet; an algebra identity holding for such a representation, holds for the abstract algebra in general: the combinatorics of the CBH expansion only involves commutators in the exponent, and is therefore identical for all irreps!

As a consequence, the algebra factor of your expression $( e^{-\sigma K_+} K_{-} e^{\sigma K_+}- K_{-} ) $ is trivial to evaluate with the doublet rep, κ, $$ \kappa_+= i\begin{pmatrix}0 &1\\ 0&0 \end{pmatrix}, ~~ \kappa_-= \kappa_+^T, ~~ \kappa_0= {1\over 2} \begin{pmatrix}1 &0\\ 0&-1 \end{pmatrix}, $$ since, explicitly, these matrices yield $$ e^{-\sigma \kappa_+} \kappa_{-} e^{\sigma \kappa_+}- \kappa_{-} =2\sigma \kappa_0+ \sigma^2 \kappa_+ ~~~\leadsto \\ e^{-\sigma K_+} K_{-} e^{\sigma K_+}- K_{-}= 2\sigma K_0+ \sigma^2 K_+ . $$

You may proceed with parsing out and evaluating more complicated expressions. Remember, the representations are only faithful for Lie algebra equations/identities, but not universal enveloping algebra such: You crucially need to know the expression evaluated above (in parenthesis) is in the Lie algebra! This does not obtain for the entire expression involving the group element.

Cosmas Zachos
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