The partition function and Wick rotation

Posted in Math, Quantum by Mike Stay on 2007 September 6

I was trying to understand Wick rotation by applying it in the case of a finite-dimensional Hilbert space, and came up with something strange. The way I’ve worked it out, it seems to map classical observables to quantum states! I’ve never heard anything like that before.

Say we have an $n$ -qubit Hilbert space $X=(\mathbb{C}^2)^{\otimes n}$ . This has the set of $n$ -bit binary strings $|0\ldots 00\rangle,$ $|0\ldots 01\rangle,$ $|0\ldots 10\rangle, \ldots$ $|1\ldots 11\rangle$ as a basis. For brevity’s sake, I’ll write these as $|0\rangle, |1\rangle, \ldots |2^n-1\rangle.$ Let

$\displaystyle H = \sum_{j=0}^{2^n-1} E_j |j\rangle\langle j|,$

where the $E_j$ are real.

Now define the operators

$\displaystyle V=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\1&-1\end{array}\right)$

and $W=V^{\otimes n}$ . $V$ is the discrete Fourier transform on a 2-dimensional space; $W$ is called the Walsh-Hadamard transform. $W$ defines a conjugate basis to the qubit basis. The most important property of $W$ for my purposes here is the fact that

$\displaystyle |\phi\rangle := W|0\rangle = N_\phi\sum_{j=0}^{2^n-1} |j\rangle,$

where the normalizing factor $N_\phi = 2^{-n/2}.$ Modulo the normalizing factor, this is a sum over all possible states.

What’s the probability amplitude that when you start in the state $|\phi\rangle$ and evolve according to $H$ , the system will still be in the state $|\phi\rangle ?$
$\displaystyle \langle \phi|e^{-iHt}|\phi\rangle = \langle 0|W \; e^{-iHt} \; W|0\rangle = N_\phi^2 \sum_{j=0}^{2^n-1} \langle j| \; e^{-iHt} \; |j\rangle$ $\displaystyle = N_\phi^2 \sum_{j=0}^{2^n-1} e^{-iE_jt}.$

Except for a factor of $i$ in the exponent and some normalization, this is the partition function for $H$ . It’s been “Wick rotated.”

What’s the probability amplitude that when you start in the state $|\phi\rangle$ and evolve according to H, the system will move to the arbitrary state $\displaystyle |\psi\rangle ?$ Well,
$\displaystyle |\psi\rangle = N_\psi \sum_{j=0}^{2^n-1} \psi_j |j\rangle,$

where each $\psi_j$ is an arbitrary complex number and $N_\psi$ is a normalizing factor. So

$\displaystyle \langle \psi| \; e^{-iHt} \; |\phi\rangle = N_\psi N_\phi \sum_{j=0}^{2^n-1} \langle j|\psi_j \; e^{-iHt} \; |j\rangle$ $\displaystyle = N_\psi N_\phi \sum_{j=0}^{2^n-1} \psi_j e^{-iE_jt}.$

If we divide this by the quantity above, we get the expectation value of a classical observable $\psi$ at “temperature” $1/it:$

$\displaystyle \frac{N_\psi N_\phi \sum\limits_{j=0}^{2^n-1} \psi_j e^{-iE_jt}}{N_\phi N_\phi \sum\limits_{j=0}^{2^n-1} e^{-iE_jt}} = \frac{N_\psi}{N_\phi} \langle\psi\rangle.$

This mapping from classical to quantum is not quantization. That maps classical observables to Hermetian operators, not to states—although, one might hit the state with the “Currying” isomorphism between states and linear transformations and get something useful.

I’m trying to work out how to connect this to a sum over paths instead of a sum over states; there’s some interesting stuff there, but I haven’t grokked it yet.

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reperiendi said, on 2007 November 29 at 5:10 pm

It has to do with the analogy between statics in n space dimensions and dynamics in (n-1) space and one time dimension. See “A spring in imaginary time.”

Each basis state describes a path in space of a spring parameterized by s; there’s an associated energy to the path. The zeta function tells what the probability is of finding the spring in that state given the spring’s temperature. A spring in equilibrium is in a minimal (really a critical) energy state.

Changing s to it changes the length parameterization to imaginary time and the static problem to a dynamic one. E -> -iS. Then, by summing exp(iS) over all paths, you get the path of least (or critical) action.

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