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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?

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  • $\begingroup$ 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! $\endgroup$ – Cosmas Zachos Sep 11 at 14:34
  • $\begingroup$ 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. $\endgroup$ – A quarky name Sep 13 at 10:34
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It is a straightforward evaluation evincing the power, charm and efficiency of matrix mechanics; I have no idea why the author feels inclined to work in the coordinate representation — let me know if you did. In any case, everything I do below concerns operators, so I drop all the superfluous carets: they are implied to hat everything here!

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) $ .

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