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Physics LibreTexts

3.9: Appendix- Some Exponential Operator Algebra

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Suppose that the commutator of two operators A,B

[A,B]=c,

where c commutes with A and B, usually it’s just a number, for instance 1 or i.

Then

[A,eλB]=[A,1+λB+(λ22!)B2+(λ33!)B3+]=λc+(λ22!)2Bc+(λ33!)3B2c+=λceλB.

That is to say, the commutator of A with eλB is proportional to eλB itself. That is reminiscent of the simple harmonic oscillator commutation relation [H,a]=ωa which led directly to the ladder of eigenvalues of H separated by ω. Will there be a similar “ladder” of eigenstates of A in general?

Assuming A (which is a general operator) has an eigenstate |a with eigenvalue a,

A|a=a|a.

Applying [A,eλB]=λceλB to the eigenstate |a:

AeλB|a=eλBA|a+λceλB|a=(a+λc)|a.

Therefore, unless it is identically zero, eλB|a is also an eigenstate of A, with eigenvalue a+λc. We conclude that instead of a ladder of eigenstates, we can apparently generate a whole continuum of eigenstates, since λ can be set arbitrarily!

To find more operator identities, premultiply [A,eλB]=λceλB by eλB to find:

eλBAeλB=A+λ[A,B]=A+λc.

This identity is only true for operators A,B whose commutator c is a number. (Well, c could be an operator, provided it still commutes with both A and B ).

Our next task is to establish the following very handy identity, which is also only true if [A,B] commutes with A and B:

eA+B=eAeBe12[A,B].

The proof is as follows:

Proof

Take f(x)=eAxeBx,

dfdx=AeAxeBx+eAxeBxB=f(x)(eBxAeBx+B)=f(x)(A+x[A,B]+B).

It is easy to check that the solution to this first-order differential equation equal to one at x=0 is

f(x)=ex(A+B)e12x2[A,B]

so taking x=1 gives the required identity,

eA+B=eAeBe12[A,B].

It also follows that eBeA=eAeBe[A,B] provided—as always—that [A,B] commutes with A and B.


This page titled 3.9: Appendix- Some Exponential Operator Algebra is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler via source content that was edited to the style and standards of the LibreTexts platform.

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