Hamiltonian for Quantum Fourier Transform 2

Sat 31 March 2018
Physics
#8.370.2x

The quantum Fourier transform is a unitary matrix $F$ that satisfies the following properties:

$\begin{align*} F^{2} &= S, & F^{3} &= F^{\dagger} = F S, & F^{4} &= S^{2} = I \end{align*}$

Here $S$ is a particular (hermitian) permutation matrix. Let

$\begin{align*} F &= \exp{(i G)}, & G^{\dagger} = G \end{align*}$

Here $G$ would be the hermitian Hamiltonian for the quantum Fourier transform. We are going to assume that $G$ has a particular form:

$\begin{equation*} G = w I + x (F + F^{\dagger}) + iy (F - F^{\dagger}) + z S \end{equation*}$

Here $w$ , $x$ , $y$ , and $z$ are real numbers. You can re-write $G$ as

$\begin{align*} G &= w I + u F + v F^{\dagger} + z S, & u &\equiv x + i y, & v &\equiv x - i y \end{align*}$

Note that

$\begin{align*} [F, S] &= 0, & [F^{\dagger}, S] &= 0, & [F, F^{\dagger}] &= 0 \end{align*}$

Thus,

$\begin{equation*} \exp{(iG)} = \exp{(iw)} \exp{(iuF)} \exp{(ivF^{\dagger})} \exp{(izS)} \end{equation*}$

Each of these factors can be expanded using trigonometric and quartic functions:

$\begin{align*} \exp{(i u F)} &= I p_{0}(u) + i F p_{1}(u) - S p_{2}(u) - i F^{\dagger} p_{3}(u) \\ \exp{(i v F^{\dagger})} &= I p_{0}(v) + i F^{\dagger} p_{1}(v) - S p_{2}(v) - i F p_{3}(v) \\ \exp{(i z S)} &= I \cos{(z)} + i S \sin{(z)} \end{align*}$

Let

$\begin{equation*} \exp{(iuF)} \exp{(ivF^{\dagger})} \exp{(izS)} = I a + F b + F^{\dagger}c + S d \end{equation*}$

In order for $G$ to be the hermitian Hamiltonian, we must have

$\begin{align*} a &= 0 & b &\neq 0 & c &= 0 & d &= 0 \end{align*}$

The $I$ coefficient will be

$\begin{multline*} a = \cos{(z)} \left[p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \right] \\ - i \sin{(z)} \left[p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \right] \end{multline*}$

This coefficient must vanish. The $F$ coefficient will be

$\begin{multline*} b = -i\cos{(z)} \left[p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \right] \\ - \sin{(z)} \left[p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{0}(v) \right] \end{multline*}$

This coefficient cannot vanish. The $F^{\dagger}$ coefficient will be

$\begin{multline*} c = i \cos{(z)} \left[p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{4}(v) \right] \\ + \sin{(z)} \left[p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \right] \end{multline*}$

This coefficient must vanish. The $S$ coefficient will be

$\begin{multline*} d = -\cos{(z)} \left[p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \right] \\ + i \sin{(z)} \left[p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \right] \end{multline*}$

This coefficient must vanish. These coefficients can be written in a simpler form:

$\begin{align*} a &= \alpha(u, v) \cos{(z)} - i \beta(u, v) \sin{(z)} & b &= -i\gamma(u, v) \cos{(z)} - \delta(u, v) \sin{(z)} \\ c &= i \delta(u, v) \cos{(z)} + \gamma(u, v) \sin{(z)} & d &= -\beta(u, v) \cos{(z)} + i \alpha(u, v) \sin{(z)} \end{align*}$

with

$\begin{align*} \alpha(u, v) &\equiv p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \\ \beta(u, v) &\equiv p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \\ \gamma(u, v) &\equiv p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \\ \delta(u, v) &\equiv p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{0}(v) \end{align*}$

Actually, these are much simpler:

$\begin{align*} 2\alpha(u, v) &= p_{0}(u - v) + p_{2}(u - v) + p_{0}(u + v) - p_{2}(u + v) \\ 2\beta(u, v) &= p_{0}(u - v) + p_{2}(u - v) - p_{0}(u + v) + p_{2}(u + v) \\ 2\gamma(u, v) &= -p_{3}(u - v) - p_{1}(u - v) + p_{3}(u + v) - p_{1}(u + v) \\ 2\delta(u, v) &= -p_{3}(u - v) - p_{1}(u - v) - p_{3}(u + v) + p_{1}(u + v) \end{align*}$

Indeed, using the definitions of the $p$ -functions gives

$\begin{align*} 2\alpha(u, v) &= \cosh{(u-v)} + \cos{(u+v)} = \cos{(2y)} + \cos{(2x)} \\ 2\beta(u, v) &= \cosh{(u-v)} - \cos{(u+v)} = \cos{(2y)} - \cos{(2x)} \\ 2\gamma(u, v) &= -\sinh{(u - v)} - \sin{(u + v)} = -i\sin{(2y)} - \sin{(2x)} \\ 2\delta(u, v) &= -\sinh{(u - v)} + \sin{(u + v)} = -i\sin{(2y)} + \sin{(2x)} \end{align*}$

Thus, $\alpha$ and $\beta$ are real and $\gamma$ and $\delta$ are complex. Since $\alpha$ and $\beta$ are real, and $a$ and $d$ must vanish, we must have

$\begin{equation*} \alpha = \beta = 0 \end{equation*}$

This can be achieved when

$\begin{align*} x_{m} &= \frac{(2m + 1) \pi}{4} & y_{n} &= \frac{(2n + 1) \pi}{4} \end{align*}$

Note that

$\begin{align*} \sin{(2x_{m})} &= (-1)^{m} & \sin{(2y_{n})} &= (-1)^{n} \end{align*}$

and thus

$\begin{align*} \gamma_{mn} &= -\frac{(-1)^{m} + i (-1)^{n}}{2} & \delta_{mn} &= \frac{(-1)^{m} - i (-1)^{n}}{2} \end{align*}$

Hence

$\begin{align*} b_{mn} &= \frac{1}{2} \cos{(z)} \left[i(-1)^{m} - (-1)^{n} \right] + \frac{1}{2} \sin{(z)} \left[i(-1)^{n} - (-1)^{m} \right] \\ c_{mn} &= \frac{1}{2} \cos{(z)} \left[i(-1)^{m} + (-1)^{n} \right] - \frac{1}{2} \sin{(z)} \left[i(-1)^{n} + (-1)^{m} \right] \end{align*}$

There are four distinct possibilities:

$\begin{align*} b_{00} &= (i - 1) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] & c_{00} &= (i + 1) \left[ \frac{\cos{(z)} - \sin{(z)}}{2} \right] \\ b_{01} &= (i + 1) \left[ \frac{\cos{(z)} - \sin{(z)}}{2} \right] & c_{01} &= (i - 1) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] \\ b_{10} &= (1 + i) \left[ \frac{\sin{(z)} - \cos{(z)}}{2} \right] & c_{10} &= (1 - i) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] \\ b_{11} &= (1 - i) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] & c_{11} &= (1 + i) \left[ \frac{\sin{(z)} - \cos{(z)}}{2} \right] \end{align*}$

Note that $c_{00} = 0$ or $c_{11} = 0$ whenever

$\begin{equation*} z_{k} = \frac{(4k + 1) \pi}{4} \end{equation*}$

Then

$\begin{align*} b_{00k} &= (-1)^{k} \left( \frac{-1 + i}{\sqrt{2}} \right) = \exp{\left(\frac{3 \pi i}{4} + k \pi i \right)} \\ b_{11k} &= (-1)^{k} \left( \frac{1 - i}{\sqrt{2}} \right) = \exp{\left(-\frac{\pi i}{4} + k \pi i \right)} \end{align*}$

Note that

$\begin{align*} b_{000} &= b_{111} = \exp{\left(\frac{3 \pi i}{4} \right)} & b_{001} &= b_{110} = \exp{\left(-\frac{\pi i}{4} \right)} \end{align*}$

In a similar way, $c_{01} = 0$ or $c_{10} = 0$ whenever

$\begin{equation*} z_{l} = \frac{(4l + 3) \pi}{4} \end{equation*}$

Then

$\begin{align*} b_{01l} &= (-1)^{l} \left( \frac{-1 - i}{\sqrt{2}} \right) = \exp{\left(-\frac{3 \pi i}{4} + l \pi i \right)} \\ b_{10l} &= (-1)^{l} \left( \frac{1 + i}{\sqrt{2}} \right) = \exp{\left(\frac{\pi i}{4} + l \pi i \right)} \end{align*}$

Note that

$\begin{align*} b_{010} &= b_{101} = \exp{\left(\frac{5 \pi i}{4} \right)} & b_{011} &= b_{100} = \exp{\left(\frac{\pi i}{4} \right)} \end{align*}$

Thus, there are four possible outcomes for the hermitian Hamiltonian:

$\begin{align*} G_{000} &= -\frac{3 \pi}{4}I + \frac{\pi}{4}(1 + i) F + \frac{\pi}{4}(1 - i) F^{\dagger} + \frac{\pi}{4} S \\ G_{001} &= \frac{\pi}{4}I + \frac{\pi}{4}(1 + i) F + \frac{\pi}{4}(1 - i) F^{\dagger} - \frac{3 \pi}{4} S \\ G_{010} &= \frac{3 \pi}{4}I + \frac{\pi}{4}(1 - i) F + \frac{\pi}{4}(1 + i) F^{\dagger} + \frac{3 \pi}{4} S \\ G_{011} &= -\frac{\pi}{4}I + \frac{\pi}{4}(1 - i) F + \frac{\pi}{4}(1 + i) F^{\dagger} - \frac{\pi}{4} S \end{align*}$

I have checked that these four hermitian Hamiltonians work for the 3-dimensional and 4-dimensional discrete Fourier transform. Indeed, my result works for any unitary matrix whose quartic power is the identity.