Witten’s Problems, Set Five — N^{o}. 2

(solution written by D. Freed)

(a) We should compute Tr e^{−T H} for

\frac{1}{2m} (−\frac{∂^{2}}{∂x^{2}} + m^{2}ω^{2}x^{2}) .

First, we can compute directly in the Hamiltonian picture. Introduce

A = \frac{1}{\sqrt{2m}} (\frac{∂}{∂x} + mωx)
A^{∗} = \frac{1}{\sqrt{2m}} (−\frac{∂}{∂x} + mωx)

Then H = A^{∗}A + \frac{ω}{2} and [A, A^{∗}] = ω. So the spectrum of H is

Spec H = \{\frac{ω}{2} + nω\}_{n∈ℤ^{≥ 0}}

and

\begin{aligned}Tr e^{−T H} & = e^{−ωT/2} ∑_{n=0}^{∞} e^{−nωT} \\ & = e^{−ωT/2} \frac{1}{1 − e^{−ωT}} \\ & = \frac{1}{2 sinh^{ωT/2}} \end{aligned} (1)

To compute the heat kernel trace as a path integral we use the Feynman-Kac formula Tr e^{−T H} = ∫ e^{−L(X)}dX, (2) where X: S^{1}(T) → ℝ is a function on the circle S^{1}(T) = ℝ/T ℤ, and

L(X) = \frac{m}{2} ∫_{0}^{T} \{Ẋ(t)^{2} + ω^{2}X(t)^{2}\} dt

Since X is periodic we can integrate by parts to obtain

(3)  L(X) = (QX, X)/2
(4)  Q = m (−\frac{∂^{2}}{∂t^{2}} + ω^{2})

where the inner product in (3) is the L^{2} inner product on S^{1}(T). So the path integral (2) is formally

Tr e^{−T H} = \frac{1}{\sqrt{det Q}} (5)

Now Q has eigenfunctions \{e^{2πint/T} \}_{n∈ℤ} with corresponding eigenvalues \{m (\frac{4π^{2}n^{2}}{T^{2}} + ω^{2})\}_{n∈ℤ}. Therefore, the determinant is the divergent product

det Q = ∏_{n∈ℤ}m (\frac{4π^{2}n^{2}}{T^{2}} + ω^{2})

= ∏_{n∈ℤ^{+}} (\frac{4π^{2}m}{T^{2}})^{2} ∏_{n=1}^{∞} (1 + (\frac{ωT}{2πn})^{2}) · mω^{2}

= ∏_{n∈ℤ^{+}} (\frac{4π^{2}m}{T^{2}}n^{2})^{2} · sinh^{2} (\frac{ωT}{2} )_{(\frac{ωT}{2} )2} · mω^{2} .

We evaluate the infinite constants using the Riemann ζ-function:

∏_{n∈ℤ^{+}} (\frac{4π^{2}m}{T^{2}} n^{2})^{2} = (\frac{4π^{2}m}{T^{2}})^{2ζ(0)} e^{−4ζ′(0)}

= \frac{T^{2}}{4π^{2}m} 4π^{2} ,

since ζ(0) = −1/2 and ζ′(0) = −\frac{1}{2} ln 2π. Altogether we have

det Q = (2 sinh \frac{ωT}{2} )^{2} ,

and plugging in (5) we obtain agreement with (1).

(b) We first evaluate the path integral. We must diagonalize (4) with antiperiodic boundary conditions. So the eigenfunctions are {e^{2πint/T} }_{n∈ℤ+1/2} with corresponding eigenvalues \{m (\frac{4π^{2}n^{2}}{T^{2}} + ω^{2})\}_{n∈ℤ+1/2}. So

\begin{aligned}det Q & = ∏_{n∈ℤ+1/2}m (\frac{4π^{2}n^{2}}{T^{2}} + ω^{2}) \\ & = ∏_{n∈ℤ+1/2}(\frac{4π^{2}m}{T^{2}} n^{2}) ∏_{n∈ℤ+1/2}(1 + (\frac{ωT}{2πn})^{2})^{2} \\ & = 4 cosh^{2} \frac{ωT}{2} , \end{aligned} (6)

from which the path integral is

\frac{1}{\sqrt{det Q}} = \frac{1}{2 cosh \frac{ωT}{2}} . (7)

In (6) we use the fact that the ζ-normalized number of integers is zero to compute

∏_{n∈ℤ+1/2}(\frac{4π^{2}m}{T^{2}} n^{2}) = ∏_{n∈ℤ+1/2}n^{2} = ∏_{n odd} n^{2} = \frac{∏_{n≠0} n^{2}}{∏_{\begin{matrix}n even \\ n≠0 \end{matrix}} n^{2}} = \frac{∏_{n≠0} n^{2}}{∏_{n≠0} 4n^{2}} = 4 ,

since the ζ-normalized number of nonzero integers is −1.

On the Hilbert space the operator J is the action on functions induced by x → −x. Notice that J anticommutes with A^{∗}, so that its action on the n^{th} eigenfunction \frac{(A^{∗})^{n}}{n!} Ω is multiplication by (−1)^{n}. Hence

Tr Je^{−T H} = e^{−ωT/2} ∑_{n=0}^{∞}(−1)^{n}e^{−nωT}

= e^{−ωT/2} \frac{1}{1 + e^{−ωT}}

= \frac{1}{2 cosh ωT/2}

This agrees with (7).