Entangling qubits

The rules of quantum mechanics have existed in their modern form since 1927 but  it was not until the 1980s that the possibility of quantum computing was first suggested. This was 40 years after the appearance of digital computers. Quantum computing is not an obvious idea if history is anything to go by.

Unlike the bits of digital computing, qubit states of a quantum computing (QC) can be rotated continuously (by probability-preserving unitary transformations). A typical quantum algorithm runs in three steps: (1) a register of N qubits is prepared in some known initial state (2) qubits are manipulated through a sequence of physical operations (corresponding to unitary transformations on multi-qubit states) (3) a probabilistic snapshot of the final quantum state is extracted (measurement). The algorithm runs in step (2). Step (3) is the read-out of results.

The list of elementary examples of QC is a short one. However, a quantum experiment often involves the same three steps above, and the result of the experiment always “computes” some quantity. Any quantum experiment has a QC interpretation.

Here I describe the quantum mechanics of N indistinguishable photons entering a beam-splitter. A pair of glass prisms glued together with a thin layer of resin form a device which partly reflects and partly transmits a beam of light. In classical optics a beam-splitter is described by complex transmission (t,t^\prime) and reflection (r,r^\prime) amplitudes that account for phase shifts of beams approaching from the left and right (unprimed and primed respectively). These amplitudes form a unitary matrix U which has the general form

    \[ U=\begin{bmatrix} t^\prime & r\\r^\prime & t\end{bmatrix} = \begin{bmatrix} \cos(\alpha) e^{i \gamma}&\sin(\alpha) e^{i \phi}\\ -\sin(\alpha) e^{-i \phi}&\cos(\alpha) e^{-i \gamma}\end{bmatrix} \]

It turns out that all of the results discussed here are independent of phases \gamma and \phi. For definiteness choose  \gamma=0 and \phi=\frac{\pi}{2}. Then

    \[ U= \begin{bmatrix} \cos(\alpha) & i\sin(\alpha) \\ i\sin(\alpha) &\cos(\alpha) \end{bmatrix} \]

If the angle \alpha=\pi/4, U=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & i \\ i &1 \end{bmatrix}. This is called “the standard 50:50 beamsplitter”. Here I allow \alpha to be different from \pi/4.

The unitary matrix U of classical optics carries over to the quantum case. Suppose that photons arrive at the beam splitter from orthogonal left or right channels as in Fig 1. The two orthogonal states are qubits denoted \left| L \rangle and \left| R \rangle. If a photo passes through the beam-splitter from the left, the photon state is rotated to U \left| L \rangle = \cos(\alpha) e^{i \gamma} \left| L \rangle + \sin(\alpha) e^{i \phi} \left| R \rangle. A photon in right channel is rotated to U \left| R \rangle = \cos(\alpha) e^{-i \gamma} \left| L \rangle - \sin(\alpha) e^{-i \phi} \left| R \rangle The transition amplitudes are \langle L |U | L \rangle = \cos(\alpha) e^{i \gamma} and \langle R |U | L \rangle= \sin(\alpha) e^{i \phi}. According to the rules of quantum mechanics the probability that a photon which entered from the left is detected in the left channel is the square modulus of the transition amplitude \left| \langle L |U | L \rangle\right|^2 = \cos^2\alpha. The probability of finding the photon in right channel is\left| \langle R |U | L \rangle\right|^2  = \sin^2\alpha = 1- \cos^2 \alpha. If \alpha = \pi/4 the probability is 50:50 as expected.

Fig 1

When N photons are present, it is best to use many-photon Fock state notation.  | n_L,n_R \rangle denotes a state with n_L photons entering in the left channel and n_R photons in the right channel (for example, the single photon state \left| R \rangle above would be represented by | 0,1 \rangle). In general the beam-splitter transforms an initial Fock state | n_L,n_R \rangle into an entangled Fock state \mathcal{U}| n_L,n_R \rangle. In general this is a superposition of all N+1 possible states with m_L photons in left and m_R right channels (satisfying m_L+m_R =N) . Of course measurement (i.e. photon counting) will yield definite values m_L, m_R. In the language of QM, measurement is said to “collapse” the state. However, by repeating the experiment many times, the full probability distribution P(m_L | n_L) can be found.

It is straightforward to calculate the general form of the beam-splitter transition amplitude \langle m_L,m_R| \mathcal{U}| n_L,n_R\rangle (the quantum mechanical calculation takes the indistinguishable boson nature of photons into account. For details with \alpha = \pi/4  see reference [1] below). The answer is:

(1)   \begin{equation*} \begin{split} \langle m_L,m_R| \mathcal{U}| n_L,n_R\rangle  & = e^{i\chi} \frac{ \sqrt{n_L! n_R!}}{\sqrt{m_L! m_R!}} \times \\  \sum_{p=\mathrm{max}(0,n_L-m_R)}^{\mathrm{min} (m_L,n_L)} & (-1)^p \left\{ \cos\alpha \right\}^{m_R+2p-n_L}\left\{ \sin \alpha\right\}^{m_L+n_L-2p} {m_L \choose p} {  m_R \choose {n_L-p  }} \end{split} \end{equation*}


Suppose that all photons enter in the same channel. In this case only a single term (p=m_L) survives in the sum in expression (1). The amplitude \langle m_R,m_L| \mathcal{U}|N,0\rangle is  e^{i \chi}(-1)^{m_L}\sqrt{\frac{N!}{m_L! (N-m_L)!}} \cos^{m_L} \alpha \sin^{N-m_L} \alpha. The square modulus of this amplitude can easily be recognised as just the binomial distribution with \cos^2 \alpha playing the role of probability.  In this case the beam-splitter is analogous to a classical analog computer, the Galton Board.

Even though the computational details look very different, this is once again the expected result from classical probability.

What happens if photons enter in both left and right channels? Suppose that one photon enters each channel. Two terms survive in expression (1) but with opposite sign and the amplitude \langle 1,1| \mathcal{U}|1,1\rangle \propto \sin^2 \alpha - \cos^2 \alpha. This vanishes for a 50:50 beam-splitter. In other words the photons enter separately but always leave together. This effect, which appears to contradict intuition based on classical probability theory, is the experimentally observed Hong-Ou-Mandel effect in quantum optics. Generalising to the case where N-1 photons enter on the left and a single photon enters on the right, once again only two terms survive in expression (1) p=m_L-1 and p=m_L, and the probability distribution is:

    \[\left| \langle m_L,m_R| \mathcal{U}| N-1,1\rangle \right|^2 = {N \choose m_L} \mathrm{q}^{m_L} (1-\mathrm{q})^{N-m_L}\frac{1}{N\mathrm{q}(1-\mathrm{q})} \left\{  m_L -\mathrm{q} N \right\}^2\]

where \mathrm{q}=\cos^2 \alpha. This is very different from the binomial distribution because it is bimodal. For example, for \mathrm{q}=1/2 and even N, it vanishes when m_L=\frac{N}{2} which is where the binomial has its maximum. This is true even when N \gg 1.

What about the general case? The probability of observing m_L photons in the left channel given that n_L photons enter the left channel (P\left(m_L | n_L\right)) are shown below for a 200-qubit beamsplitter and \alpha = \frac{3}{16} \pi:

Another representation of the probability distribution (P(m_L | n_L)) is shown below:

The N-photon beam-splitter experiment illustrates the preparation, transformation, measurement steps of QC and “computes” the square modulus of expression (1). Note that the factorials appearing in expression (1) are problematic on a classical computer when N > 200 without resorting to Stirling’s approximation.

Generalisations of the beam-splitter experiment to the multi-channel case known as Boson Sampling [2] have been proposed as a means to demonstrate quantum advantage, a computation that can be executed in QC but is infeasible on a classical computer. The quantity computed in Boson Sampling is closely related to the permanent of a matrix, a computationally hard problem on a classical computer.



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