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?

  • $\begingroup$ Have a look at this experiment phys.org/news/…. It says that once the situation changes for slit A with a filter the solution changes, the filter at the slit becoming a source for the photons that interact with it, destroying the coherence. Your inclusion of systemA at Ca does something similar. You cannot have one wavefunction imo, you will need a density matrix formalism. $\endgroup$
    – anna v
    Feb 2, 2016 at 4:44
  • $\begingroup$ @anna_v: After time $t_{A+\tau}$, I think that the particle alone (not including system A), will need to be described by a reduced density matrix (of the composite density matrix) and the particle will be in a mixed state. Is that what you mean? $\endgroup$
    – David
    Feb 2, 2016 at 5:01
  • $\begingroup$ yes , at least that is what the link I gave implies. The solution which will give the probability of finding the particle in a or b will be mixed. the particle is whole each time it is measured $\endgroup$
    – anna v
    Feb 2, 2016 at 5:09
  • $\begingroup$ @anna v: But I think that the composite system, of the particle and system A, will have a wave function in a pure state, as I described above, where its two terms, for the two different channels, have different interaction Hamiltonians for how the particle and system A interact. $\endgroup$
    – David
    Feb 2, 2016 at 5:13
  • $\begingroup$ This paper helps motivate my thinking: quantum.phys.cmu.edu/CQT/chaps/cqt18.pdf $\endgroup$
    – David
    Feb 2, 2016 at 5:14

1 Answer 1


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.

*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).

  • $\begingroup$ Mark Mitchison: (1) of (2): I think $$ \langle \Psi_{P_i}\rvert \langle \Psi_{A_i}\rvert U(t) \lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle $$ could also be derived by noticing that $$\lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle $$ is the initial state vector of the composite system of system A and the particle (before interaction), $$U(t) \lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle $$ is the state vector of the composite system at time t (sometime after interaction). $\endgroup$
    – David
    Feb 6, 2016 at 19:14
  • $\begingroup$ (2) of (2): So, $$ \langle \Psi_{P_i}\rvert \langle \Psi_{A_i}\rvert U(t) \lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle $$ is a measure of how different the composite system is after interaction, from what it was before the interaction. If there is no difference, the value of the inner product is 1 and we would expect maximum interference. if the value is zero, then those vectors have no overlap and so no interference. Does that seem like another way to think of it? $\endgroup$
    – David
    Feb 6, 2016 at 19:16
  • $\begingroup$ What is interesting about this is that a quantum particle's wave function can live out one scenario in channel a, even causing back-action to System A there, but live out a totally different scenario in channel b. If the particle is detected in channel b, then the interaction with system A, and subsequent back-action on system A is nullified. Of course, that is only in the, perhaps, rare case where a wave function is split into two non-zero packets, each going in a different direction and interacting with different highly localized things. It kind of captures my imagination. $\endgroup$
    – David
    Feb 6, 2016 at 19:24
  • $\begingroup$ @David Indeed, the visibility is just given by the overlap between the initial and final states of A and P. I agree, interferometry is a very rich and fascinating subject! $\endgroup$ Feb 7, 2016 at 4:54
  • $\begingroup$ @ Mark Mitchison: One last thing, I am not sure how derived: \begin{align}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}, given \begin{align} U(t) = T \exp \left ( \int_0^t\mathrm{d}s \,H_{AP}(s)\otimes \lvert C_a \rangle \langle C_a\rvert \right), \end{align} $\endgroup$
    – David
    Feb 7, 2016 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.