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# Appendix: Some Exponential Operator Algebra

Suppose that the commutator of two operators $$A$$,$$B$$

$[A,B]=c, \label{3.6.50}$

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

Then

\begin{align} [A,e^{\lambda B}] &= \left[A,1+\lambda B+ \left(\dfrac{\lambda^2}{2!} \right)B^2+ \left(\dfrac{\lambda^3}{3!}\right)B^3+\dots\right] \\[5pt] &= \lambda c+ \left(\dfrac{\lambda^2}{2!}\right)^2Bc+ \left(\dfrac{\lambda^3}{3!}\right)^3B^2c+\dots \\[5pt] &= \lambda ce^{\lambda B}. \label{3.6.51} \end{align}

That is to say, the commutator of $$A$$ with $$e^{\lambda B}$$ is proportional to $$e^{\lambda B}$$ itself. That is reminiscent of the simple harmonic oscillator commutation relation $$[H,a^{\dagger}]=\hbar\omega a^{\dagger}$$ which led directly to the ladder of eigenvalues of $$H$$ separated by $$\hbar\omega$$. Will there be a similar “ladder” of eigenstates of $$A$$ in general?

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

$A|a\rangle=a|a\rangle. \label{3.6.52}$

Applying $$[A,e^{\lambda B}]=\lambda ce^{\lambda B}$$ to the eigenstate $$|a\rangle$$:

$Ae^{\lambda B}|a\rangle=e^{\lambda B}A|a\rangle+\lambda ce^{\lambda B}|a\rangle=(a+\lambda c)|a\rangle. \label{3.6.53}$

Therefore, unless it is identically zero, $$e^{\lambda B}|a\rangle$$ is also an eigenstate of $$A$$, with eigenvalue $$a+\lambda c$$. We conclude that instead of a ladder of eigenstates, we can apparently generate a whole continuum of eigenstates, since $$\lambda$$ can be set arbitrarily!

To find more operator identities, premultiply $$[A,e^{\lambda B}]=\lambda ce^{\lambda B}$$ by $$e^{-\lambda B}$$ to find:

$e^{-\lambda B}Ae^{\lambda B}=A+\lambda[A,B]=A+\lambda c. \label{3.6.54}$

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 $$B$$:

$e^{A+B}=e^Ae^Be-\frac{1}{2}[A,B]. \label{3.6.55}$

Proof

The proof is as follows:

Take $$f(x)=e^{Ax}e^{Bx}$$, $df/dx=Ae^{Ax}e^{Bx}+e^{Ax}e^{Bx}B=f(x)(e^{-Bx}Ae^{Bx}+B)=f(x)(A+x[A,B]+B). \label{3.6.56}$

It is easy to check that the solution to this first-order differential equation equal to one at $$x=0$$ is $f(x)=e^{x(A+B)}e^{\frac{1}{2}x^2[A,B]} \label{3.6.57}$

so taking $$x=1$$ gives the required identity,

$e^{A+B}=e^Ae^Be^{-\frac{1}{2}[A,B]}.$

$$\square$$

It also follows that $$e^Be^A=e^Ae^Be^{-[A,B]}$$ provided—as always—that $$[A,B]$$ commutes with $$A$$ and $$B$$.