wavefunction - Single quantum particle in beam splitter, with different systems located in each channel

Suppose a quantum mechanical particle enters a beam-splitter, which sends its wave packets into two mutually orthogonal channels, $C_a$ and $C_b$. Suppose that $C_a$ also contains System A, with observable $|\Psi_A\rangle$, with which the particle's (in general, different) observable $|\Psi_P\rangle$ interacts, via Hamiltonian $H_{AP}$. $C_b$ does not contain system A, nor any other system (just free space).

Without System A, once within the beam splitter, we could use the simple description:

$$\begin{equation} |\Psi\rangle = (|C_a\rangle + |C_b\rangle) / \sqrt 2 \end{equation}$$

With System A located in $C_a$, after the particle enters the beam splitter, the wave function of the composite system evolves to:

$$\begin{equation}|\Psi\rangle_{t = t_{A+\tau}} = \exp \left( -i\int^{t_{A+\tau}}_{t_A} H_{AP} \mathrm{d}t \right)|\Psi_{A_i}\rangle|\Psi_{P_i}\rangle |C_a\rangle + |\Psi_{A_i}\rangle|\Psi_{P_i}\rangle |C_b\rangle \end{equation}$$

where the notation "$|\alpha\rangle |\beta\rangle$" represents the tensor product of $|\alpha\rangle$ and $|\beta\rangle$. $|\Psi_{A_i}\rangle$ is the initial state of System A's observable before interacting with the particle, $|\Psi_{P_i}\rangle$ is the initial state of the particle's observable before interacting with System A, $t_A$ is the time when the particle begins interacting with System A, and $t_{A+\tau}$ is when the particle stops interacting with System A. Normalization is suppressed for simplicity.

In the second equation, the $C_a$ term contains a Hamiltonian that associates specific states of the particle with specific states of System A (entanglement of the particle with system A). The $C_b$ term contains the "null" Hamiltonian, which associates each state of the particle with all states of System A (for all intents and purposes, no entanglement of the particle with System A). Further, the second equation implies that if both channels were later recombined, there would be no interference between their components of the wave function of the particle.

Does the above description sound correct?

Answer

Let's write the full Hamiltonian of the problem as

$$H = H_A + H_P + H_C + V,$$ where $H_A + H_P + H_C$ are the local Hamiltonians of $A$, $P$ and $C$, and their interaction is $V$. If you want the interaction to take place in arm $a$ of the interferometer, the interaction should be of the form $$V = H_{AP} \otimes \lvert C_a \rangle \langle C_a\rvert.$$ For an interferometer, we should have that $\lvert C_{a/b}\rangle$ are eigenstates of $H_C$, which means that $[V,H_C] = 0$. Moving to an interaction picture generated by $H_0 = H_A + H_P + H_C$, the time evolution operator is $$\begin{align} U(t) = T \exp \left ( \int_0^t\mathrm{d}s \,V(s) \right) & = T \exp \left ( \int_0^t\mathrm{d}s \,H_{AP}(s)\otimes \lvert C_a \rangle \langle C_a\rvert \right), \end{align}$$ where $V(s)$, $H_{AP}(s)$ are interaction-picture operators, e.g. $V(t) = \mathrm{e}^{\mathrm{i}H_0t}V\mathrm{e}^{-\mathrm{i}H_0t}$.

Now compute the evolution of the quantum state in the interaction picture

$$\begin{align} \lvert \Psi(t)\rangle & = U(t)\lvert \Psi(0)\rangle \\ &= U(t)\lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle \left[ \lvert C_a \rangle + \lvert C_b \rangle \right] \\ & = T \exp \left ( \int_0^t\mathrm{d}s \,H_{AP}(s) \right)\lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle\lvert C_a \rangle + \lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle\lvert C_b \rangle, \end{align}$$ where I have neglected an overall normalisation factor.

Interference indicates the presence of coherence between the path states $\lvert C_{a/b}\rangle$. In other words, if the path states are recombined at a second beam splitter (or some other kind of interference experiment) the output measures the matrix element* $\langle C_a \lvert \rho_C \rvert C_b\rangle$, where $\rho_C$ is the reduced density matrix corresponding to the path degrees of freedom, i.e.

$$\rho_C(t) = \mathrm{Tr}_{AP} \left \lbrace\lvert \Psi(t) \rangle \langle \Psi(t)\rvert \right\rbrace,$$ where $\mathrm{Tr}_{AP}$ means the partial trace over the Hilbert space of $A$ and $P$.

Putting everything together, we obtain

$$\langle C_a \lvert \rho_C(t) \rvert C_b\rangle = \langle \Psi_{P_i}\rvert \langle \Psi_{A_i}\rvert T \exp \left ( \int_0^t\mathrm{d}s \,H_{AP}(s) \right)\lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle.$$ The magnitude of this expression is a real number between 0 and 1 giving the interference visibility (0 means no interference, 1 means maximum interference). Note that there is also a phase which affects the result of the interference experiment. For example, in a Mach-Zehnder interferometer, the phase determines which output port the particle (photon) will be detected. If you instead are doing a double-slit type experiment, the phase determines the position of the interference maxima on the detecting screen.

Depending on the kind of interaction $H_{AP}$, you could get anything from a simple phase shift to a complete loss of interference. Generally speaking, if the states $\lvert \Psi_{P_i/A_i}\rangle$ are eigenstates of $H_{AP}(t)$ at all times, then you just get a phase shift. However, if the interaction leads to transitions to different states of $P,A$, then the interference will be partially or completely destroyed.

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*Notice that $\langle C_a \lvert \rho_C \rvert C_b\rangle$ is complex and therefore cannot be written as the expectation value of a Hermitian operator. Thus it is not strictly speaking a quantum mechanical observable. A given interferometer set-up can only measure either the real or the imaginary part of $\langle C_a \lvert \rho_C \rvert C_b\rangle$. (Or to be more precise, it can measure only one of the two real numbers needed to specify a single complex number: it might instead measure the magnitude or the phase, for example).