Transfer operator spectrum of the discretized harmonic oscillator

Question

I'm reading "An introduction to quantum fields on a lattice" by Jan Smit.

In chapter 2, the transfer operator 𝑇̂ is defined and shown to be equal to $$ \hat{T} = e^{-\omega^2 \hat{q}^2/4} e ^{-\hat{p}^2/2}e^{-\omega^2 \hat{q}^2/4} $$ (There is another question about this section: Eigenvalue spectrum of the transfer operator for the harmonic oscillator)

With the usual coordinate representation: $$\hat{q} \to q , ~~~ \hat{p} \to -i \partial / \partial q$$

The coordinate representation of the ground state is given by: $$ \langle q|0\rangle = e^{- \frac{1}{2} \sinh \tilde{\omega} \, q²} $$ $ \tilde{\omega}$ and $\omega$ are related via $$ \cosh \tilde{\omega} = 1 + \frac{1}{2} \omega^2$$

Now the question: I don't know how to derive this equation: $$ \hat{T} |0\rangle = e^{-\frac{1}{2} \tilde{\omega}}|0\rangle $$

What I thought of so far:

• To bring $\hat{T}$ to the form $\hat{T}=e^{-\hat{H}}$ by using the Baker-Campbell-Hausdorff formula. But the exact BCH, $$ e^{X}e^{Y}=e^{{X+Y+[X,Y]/2}} $$ is not applicable here, since $[X,[X,Y]] \neq 0$, for $X = \hat{p}^2$ and $Y = \hat{q}^2$.

• Use the coordinate representation of $\hat{T} |0\rangle $, e.g. $$ e^{-\omega^2 \hat{q}^2/4} e ^{-\hat{p}^2/2}e^{-\omega^2 \hat{q}^2/4} e^{- \frac{1}{2} \sinh \tilde{\omega} \, q²}, $$ where the last part is the ground state. Then I use $ e^{-\hat{p}^2/2} = 1 - \frac{p²}{2} + \frac{p^4}{8} + ...$ and then insert the coordinate representation of the momentum operator, giving a ladder of derivatives of zeroth, then second, then fourth order... But this also does not yield the desired result.

I am thankful for any ideas!

Answer 1

Inserting complete sets of coordinate eigenstates yields finally the result:

$$ \langle q |\hat{T} |0\rangle = \int dq^{\prime \prime} \int dq^{\prime} \langle q | e^{-\omega^2 \hat{q}^2/4} |q^{\prime} \rangle \langle q^{\prime} | e^{-\hat{p}^2/2} |q^{\prime \prime} \rangle \langle q^{\prime \prime} |e^{-\omega^2 \hat{q}^2/4} |0\rangle $$

using $$ \langle q | e^{-\omega^2 \hat{q}^2/4} |q^{\prime} \rangle= e^{-\omega^2 q^{\prime 2}/4} \delta(q-q^{\prime}) $$ $$ \langle q^{\prime\prime} | e^{-\omega^2 \hat{q}^2/4} |0 \rangle= e^{-\omega^2 q^{\prime \prime 2}/4} e^{- \frac{1}{2} \sinh \tilde{\omega} \, q^{\prime\prime 2}} $$ and $$ \langle q^\prime | e^{-\hat{p}^2/2} |q^{\prime \prime} \rangle= \sqrt{ \frac{1}{2 \pi}} e^{-\frac{(q^\prime-q^{\prime \prime})^2}{2}} $$ (can be shown by inserting complete set of momentum eigenstates). One arrives at the desired result $$ \langle q |\hat{T} |0\rangle= e^{-\frac{1}{2} \tilde{\omega}} e^{- \frac{1}{2} \sinh \tilde{\omega} \, q^{ 2}} = e^{-\frac{1}{2} \tilde{\omega}} \langle q |\hat{T} |0\rangle $$

The only thing needed are Gauss integral and the relation $\cosh \tilde{\omega} = 1 + \frac{1}{2} \omega^2$