# Appendix: Some Exponential Operator Algebra

- Page ID
- 11574

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 \(B\).

# Contributor

- Michael Fowler (Beams Professor, Department of Physics, University of Virginia)