# Eigenvalue spectrum of the transfer operator for the harmonic oscillator

I'm reading "An introduction to quantum fields on a lattice" by Jan Smit.
In chapter 2, the transfer operator $$\hat{T}$$ is defined and shown to be equal to $$\hat{T} = e^{-\omega^2 \hat{q}^2/4} e ^{-\hat{p}^2/2}e^{-\omega^2 \hat{q}^2/4}$$ for the harmonic oscillator.

There is a line in the derivation of its eigenvalue spectrum which says "Using the representation $$\hat{q} \to q , ~~~ \hat{p} \to -i \partial / \partial q$$, one obtains $$\hat{T} \left(\begin{matrix}\hat{p} \\ \hat{q} \end{matrix}\right) = M \left(\begin{matrix}\hat{p} \\ \hat{q} \end{matrix}\right) \hat{T}~~~~\hbox{where} \\ M = \left( \begin{matrix} 1+ \frac{1}{2} \omega^2 & i \\ -i(2+ \frac{1}{2} \omega^2)\frac{1}{2} \omega^2 & 1+\frac{1}{2} \omega^2 \end{matrix} \right)$$". How would one go about showing this?

• You almost certainly have reversed $\hat p$ and $\hat q$ in your vectors. It is a straightforward commutation slip. Why the author prefers to work in the coordinate representation is obscure to me as well! – Cosmas Zachos Sep 11 '19 at 14:34
• Thanks so much for your help with this! That makes sense, and is a very nice technique to know. I double checked and it would seem like the vector elements $\hat{q}$ and $\hat{p}$ are ordered as I first said but your derivation makes sense. Also, I don't seem to see the reasoning behind use of the coordinate representation. – A quarky name Sep 13 '19 at 10:34

The fundamental adjoint action Hadamard identity one always uses is $$e^{A} B e^{-A} \equiv \operatorname{Ad}_{e^A} ~ B = e^{\operatorname{ad}_A} B\equiv e^{[A,\bullet ] } B= B+ [A,B]+ \frac{1}{2} [A, [A,B]]+...$$ but (oh, the joy!) only the linear terms in A will survive in our manipulations, since the arguments thereof are always linear, by $$[q,p]=i$$.
Just compute away: $$\bbox[yellow]{ e^{-\frac{\omega^2}{4} [q^2,\bullet]} ~ q =q, \qquad e^{-\frac{\omega^2}{4} [q^2,\bullet]} ~ p = p -i\frac{\omega^2}{2} q \\ e^{-\frac{1}{2} [p^2,\bullet]}~ p = p , \qquad e^{-\frac{1}{2} [p^2,\bullet]} ~q = q + iq \qquad}~.$$
Consequently, $$T p T^{-1}= e^{-\frac{\omega^2}{4} [q^2,\bullet]} e^{-\frac{1}{2} [p^2,\bullet]}e^{-\frac{\omega^2}{4} [q^2,\bullet]} ~ p \\ \qquad \qquad = (1+\omega^2/2) p -i(\omega^2/2)(2+\omega^2/2)q ~~, \\T q T^{-1}= e^{-\frac{\omega^2}{4} [q^2,\bullet]} e^{-\frac{1}{2} [p^2,\bullet]}e^{-\frac{\omega^2}{4} [q^2,\bullet]} ~ q \\ = i p +(1+\omega^2/2)q ~~.$$
This is your target vector equation if only you reverse the order of p and q in your operator vectors. *I believe the correct equation you meant to write is actually $$\hat{T} \left(\begin{matrix}\hat{q} \\ \hat{p} \end{matrix}\right)\hat{T}^{-1} = M \left(\begin{matrix}\hat{q} \\ \hat{p} \end{matrix}\right)$$ .