Quantum Computing 2.5 -- Mirrors and beam splitters

Computer Science Level 5

Consider the following optical setup with a single photon source, two beam splitters, two mirrors and two detectors 0 and 1:

The light source now emits a single photon. What is the probability that this light particle will be measured at detector 0?

Details and assumptions: The mirrors and beam splitters consist of a glass plate coated with a dielectric film that reflects light. In the case of the beam splitter, the reflectivity and transmittity are both equal to 50%. The reflectivity of the mirror is 100%. The reflection at the air-dielectric interface leads to a phase shift by the angle π \pi . By contrast, there is no phase shift in reflection at the glass-dielectric interface and in transmission. We assume that the optical setup is lossless and that the detectors work 100% efficient.

Note: The photon may be below or above the beam splitters, so we can refer to these positions as different quantum states ∣ 0 ⟩ | 0 \rangle and ∣ 1 ⟩ | 1 \rangle . The photon starts in the state ∣ 0 ⟩ | 0 \rangle and assumes a final state ∣ 0 ⟩ | 0 \rangle , if it is measured by the detector 0. The optical setup can be considered as a 1-qubit quantum gate, which is described by an operator U U . Therefore, the wanted probability is calculated by P ∣ 0 ⟩ = ∣ ⟨ 0 ∣ U ∣ 0 ⟩ ∣ 2 P_ {| 0 \rangle} = | \langle 0 | U | 0 \rangle | ^ 2 .

Here is an older version of this problem with a more complicated setup.

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Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

We assign the state ∣ 0 ⟩ | 0 \rangle to a photon in the lower half, while a photon in the upper half has a state ∣ 1 ⟩ | 1 \rangle . The beam splitter generates from the base states ∣ 0 ⟩ | 0 \rangle and ∣ 1 ⟩ | 1 \rangle a uniform superposition α ∣ 0 ⟩ + β ∣ 1 ⟩ \alpha | 0 \rangle + \beta | 1 \rangle with the probabilities ∣ a l p h a ∣ 2 = ∣ β ∣ 2 = 1 2 | alpha | ^ 2 = | \beta | ^ 2 = \frac {1} {2} . For an incident beam in the state ∣ 1 ⟩ | 1 \rangle , the beam splitter produces a phase shift between reflected and transmitted beam with the phase factor e i π = − 1 e ^ {i \pi} = -1 . In the other case, however, there is no phase shift between the outgoing beams. The beam splitter thus acts as a Hadamard gate: ( beam splitter ) ∣ 0 ⟩ = ∣ 0 ⟩ + ∣ 1 ⟩ 2 ( beam splitter ) ∣ 1 ⟩ = ∣ 0 ⟩ − ∣ 1 ⟩ 2 ⇒ beam splitter ≡ H = 1 2 ( 1 1 1 − 1 ) \begin{aligned} (\text{beam splitter}) |0\rangle &= \frac{|0\rangle + |1\rangle}{\sqrt{2}} \\ (\text{beam splitter}) |1\rangle &= \frac{|0\rangle - |1\rangle}{\sqrt{2}} \\ \Rightarrow \quad \text{beam splitter} \equiv H &= \frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \end{aligned} At the mirrors the light is only reflected, so that the incident photon in the state ∣ 0 ⟩ | 0 \rangle or ∣ 1 ⟩ | 1 \rangle also remains in the respective base state. Since the mirrors are oriented so that the coated side points upwards, only in the state 0 ⟩ 0 \rangle a phase shift with the phase factor e i π = − 1 e ^ {i \pi} = -1 results. Thus, the two mirrors in the { 0 ⟩ , ∣ 1 ⟩ } \{0 \rangle, | 1 \rangle \} -basis correspond to a phase flip ( Z Z -gate): ( mirrors ) ∣ 0 ⟩ = − ∣ 0 ⟩ ( mirrors ) ∣ 1 ⟩ = ∣ 1 ⟩ ⇒ mirrors ≡ − Z = ( − 1 0 0 − 1 ) \begin{aligned} (\text{mirrors}) |0\rangle &= - |0\rangle \\ (\text{mirrors}) |1\rangle &= |1\rangle \\ \Rightarrow \quad \text{mirrors} \equiv -Z &= \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right) \end{aligned} The entire optical setup thus corresponds to an operation: U = ( beam splitter ) ⋅ ( mirrors ) ⋅ ( beam splitter ) = H ⋅ ( − Z ) ⋅ H = − 1 2 ( 1 1 1 − 1 ) ( 1 0 0 − 1 ) ( 1 1 1 − 1 ) = − 1 2 ( 1 1 1 − 1 ) ( 1 1 − 1 1 ) = − 1 2 ( 0 2 2 0 ) = − ( 0 1 1 0 ) = − X \begin{aligned} U &= (\text{beam splitter}) \cdot (\text{mirrors}) \cdot (\text{beam splitter}) \\ &= H \cdot (- Z) \cdot H \\ &= -\frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \\ &= -\frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right) \\ &= -\frac{1}{2} \left( \begin{array}{cc} 0 & 2 \\ 2 & 0 \end{array} \right) \\ &= -\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \\ &= - X \end{aligned} Thus, P = ∣ ⟨ 0 ∣ U ∣ 0 ⟩ ∣ 2 = ∣ ⟨ 0 ∣ X ∣ 0 ⟩ ∣ 2 = ∣ ⟨ 0 ∣ 1 ⟩ ∣ 2 = 0 P = |\langle 0 |U |0\rangle |^2 = |\langle 0 |X |0\rangle |^2 = |\langle 0 |1\rangle |^2 = 0

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