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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 \(B\). 

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