Skip to main content
Chemistry LibreTexts

260: Single Photon Interference - Mathcad version

  • Page ID
    137795
  • The schematic diagram below shows a Mach-Zehnder interferometer for photons. When the experiment is run so that there is only one photon in the apparatus at any time, the photon is always detected at D1 and never at D2.

    This surprising phenomenon will be analyzed using matrix mechanics. State vectors for photon motion in the x- and y-direction, plus matrix operators for beam splitters and mirrors are defined below. For background and references to the primary literature see: V. Scarani and A. Suarez, "Introducing Quantum Mechanics: One-particle Interferences," Am. J. Phys. 66, 718-721 (1998).

    Screen Shot 2019-02-20 at 1.45.08 PM.png

    Orthonormal basis states:

    Photon moving in x-direction: \[ x = \begin{bmatrix}
    1\\
    0
    \end{bmatrix} ~~~x^T x = 1\]

    Photon moving in y-direction: \[ y = \begin{bmatrix}
    0\\
    1
    \end{bmatrix} ~~~y^T y = 1~~~ x^T y = 0 ~~~ y^T x = 0\]

    Operators:

    Operator for interaction with the mirror:

    \[ M = \begin{bmatrix}
    0 & 1\\
    1 & 0
    \end{bmatrix}\]

    Operator for interaction with a 50/50 beam splitter:

    \[ BS = \frac{1}{ \sqrt{2}} \begin{bmatrix}
    1 & i\\
    i & 1
    \end{bmatrix}\]

    Operations:

    \[ M (x) = \begin{bmatrix}
    0\\
    1
    \end{bmatrix} ~~~ M (y) = \begin{bmatrix}
    1\\
    0
    \end{bmatrix} ~~~ BS (x) = \begin{bmatrix}
    0.707\\
    0.707i
    \end{bmatrix} ~~~ BS (y) = \begin{bmatrix}
    0.707i\\
    0.707
    \end{bmatrix} ~~~ BS (M) BS (x) = \begin{bmatrix}
    i\\
    0
    \end{bmatrix}\]

    Quantum Mechanical Calculation of Experimental Results:

    To be detected at D1 the photon must be moving in the x-direction (photon state = |x>). To be detected at D2 the photon must be moving in the y-direction (photon state = |y>). It is shown below that the probability the photon is moving in the x-direction is 1 and the probability it is moving in the y-direction is 0. In its course from source to detector the photon encounters a beam splitter, a mirror, and another beam splitter. Thus,

    Probability photon will arrive at detector D1:

    \[ (| x^T BS(M)BS(x)|)^2 = 1\]

    Probability photon will arrive at detector D2:

    \[ (| x^T BS(M)BS(x)|)^2 = 1\]

    Further analysis:

    The photon leaves the source traveling in the x-direction. Interaction with the beam splitter puts the photon in an even linear superposition of traveling in the x- and y-directions with a 90o phase shift ( \( \frac{ \pi}{2}\) or i) assigned by convention to motion in the y-direction (see note below).

    \[ BS (x) = \begin{bmatrix}
    0.707\\
    0.707i
    \end{bmatrix} ~~~ \frac{x+iy}{ \sqrt{2}} = \begin{bmatrix}
    0.707\\
    0.707i
    \end{bmatrix}\]

    The interaction of this state with the mirrors transfers the 90o phase shift to motion in the x-direction.

    \[ (M) BS (x) = \begin{bmatrix}
    0.707i\\
    0.707
    \end{bmatrix} = \frac{ix+y}{ \sqrt{2}} = \begin{bmatrix}
    0.707i\\
    0.707
    \end{bmatrix}\]

    Finally the interaction of this state with the second beam splitter yields a 90o phase-shifted photon travelling in the x-direction.

    \[ BS(M)BS(x) = \begin{bmatrix}
    i\\
    0
    \end{bmatrix} ~~~ ix = \begin{bmatrix}
    i\\
    0
    \end{bmatrix}\]

    Thus, the probability amplitude that it will be detected at D1 is i and the probability amplitude that it will be detected at D2 is 0.

    \[ x^T BS(M)BS(x) = (i) ~~~ y^T BS(M)BS(x) = (0)\]

    The probability for an event is the square of the absolute magnitude of the probability amplitude, so the probability that the photon will be detected at D1 is 1.

    Note: The justification for the 90 (\( \frac{ \pi}{2}\) or i) phase shift between transmission and reflection at the beam splitter is conservation of energy. Assuming there is no phase shift requires that the BS operator be defined as:

    \[ BS = \frac{1}{ \sqrt{2}} \begin{bmatrix}
    1 & 1\\
    1 & 1
    \end{bmatrix}\]

    This leads to the following calculation for the probability of the detection events:

    Probability photon will arrive at detector D1: \( (| x^T BS(M)BS(x)|)^2 = 1\)

    Probability photon will arrive at detector D2: \( (| y^T BS(M)BS(x)|)^2 = 1\)

    According to this analysis, the single photon leaving the source has arrived at both detectors. One photon has become two, an obvious violation of the energy conservation principle.