# Transfer operator spectrum of the discretized harmonic oscillator

Physics Asked by completeStranger on January 5, 2021

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

In chapter 2, the transfer operator ?̂ 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:
$$langle q|0rangle = e^{- frac{1}{2} sinh tilde{omega} , q²}$$
$$tilde{omega}$$ 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:
$$hat{T} |0rangle = e^{-frac{1}{2} tilde{omega}}|0rangle$$

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 $$hat{T} |0rangle$$, 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!

Inserting complete sets of coordinate eigenstates yields finally the result:

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

using $$langle q | e^{-omega^2 hat{q}^2/4} |q^{prime} rangle= e^{-omega^2 q^{prime 2}/4} delta(q-q^{prime})$$ $$langle q^{primeprime} | e^{-omega^2 hat{q}^2/4} |0 rangle= e^{-omega^2 q^{prime prime 2}/4} e^{- frac{1}{2} sinh tilde{omega} , q^{primeprime 2}}$$ 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 $$langle q |hat{T} |0rangle= e^{-frac{1}{2} tilde{omega}} e^{- frac{1}{2} sinh tilde{omega} , q^{ 2}} = e^{-frac{1}{2} tilde{omega}} langle q |hat{T} |0rangle$$

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

Correct answer by completeStranger on January 5, 2021