3.9: Appendix- Some Exponential Operator Algebra
( \newcommand{\kernel}{\mathrm{null}\,}\)
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 iℏ.
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†]=ℏωa† which led directly to the ladder of eigenvalues of H separated by ℏω. 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 e−λB to find:
e−λBAeλB=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=eAeBe−12[A,B].
The proof is as follows:
Proof
Take f(x)=eAxeBx,
dfdx=AeAxeBx+eAxeBxB=f(x)(e−BxAeBx+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=eAeBe−12[A,B].
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It also follows that eBeA=eAeBe−[A,B] provided—as always—that [A,B] commutes with A and B.
Contributor
- Michael Fowler (Beams Professor, Department of Physics, University of Virginia)