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Having read this question and answers to it, I've learned that somehow two light beams from independent sources can actually produce interference pattern, if the properties of their sources are good enough.

Now, this conflicts with my understanding of quantum particles. I suppose that interference of independent beams means interference between pairs of particles from each beam. Consider a pair of non-relativistic non-interacting bosons such that it can be described by Schrödinger equation. Its state function, up to normalization, would be $$\Psi(\vec r_1,\vec r_2,t)=\psi_1(\vec r_1,t)\psi_2(\vec r_2,t)+\psi_2(\vec r_1,t)\psi_1(\vec r_2,t).$$

Let now $\psi_1(\vec r,t)$ be a 2D gaussian wave packet going along $y=x$ axis, and $\psi_2(\vec r,t)$ be a similar wave packet going in direction of $y=-x$. Obviously there exists an area where they "intersect". Let this point be around $(x,y)=(0,0).$ At this point we might place a detector, along $y=0$, which would show us the intensity of the particle beam, which would be governed by the following formula:

$$D(x)=\int_{-\infty}^\infty |\Psi((x,0),(x_2,0),t)|^2\;dx_2.\tag1$$

Now I wasn't able to find this integral analytically, but numeric calculations show that there's no interference pattern on the screen, the intensities of both beams just add up.

This is what I get for particle density in $(x,y)$ space — evaluated as in $(1)$:

enter image description here

And this is what I'd expect based on the answers to the question mentioned above (this was generated as a probability density for single particle in two-packets state):

enter image description here

So the question: what's so special about photons that they do exhibit interference pattern, while usual non-relativistic bosons obeying Schrödinger equation don't? I suppose the core reasons might be some of:

  1. Non-relativistsness of Schrödinger equation, which my analysis was based on
  2. Different form of equations governing evolution of light, i.e. Maxwell's equations vs Schrödinger's one.
  3. Something related to QED, which is not taken into account in QM
  4. My mistake
  5. Something other

What are the real reasons for this discrepancy?

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  • $\begingroup$ "I suppose that interference of independent beams means interference between pairs of particles from each beam." That isn't quite right. In quantum mechanics each particle interferes with itself. $\endgroup$ – Chris Mueller Mar 18 '14 at 13:05
  • $\begingroup$ How about this nature.com/nature/journal/v448/n7151/abs/nature05955.html $\endgroup$ – anna v Mar 18 '14 at 13:09
  • $\begingroup$ @ChrisMueller This isn't quite the case in this setup. Here there're two beams of freely-propagating particles. The particles would interfere with themselves and produce interference pattern if there were some inhomogeneities in potential, which would scatter the particles. But in this case the only possible interference is between particles of different beams. $\endgroup$ – Ruslan Mar 18 '14 at 13:09
  • $\begingroup$ @annav that article seems to be about joint probability, not interference pattern on integrating screen. It doesn't contradict my understanding of quantum particles. What does is the interference pattern on the integrating screen. $\endgroup$ – Ruslan Mar 18 '14 at 13:12
  • $\begingroup$ Indeed. I somehow misread your statement to mean the two beams in a classic double slit experiment. $\endgroup$ – Chris Mueller Mar 18 '14 at 14:09
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My problem appears to be with the initial state of the system, which I have written as

$$\left| \Psi\right\rangle=\left|\psi_1\right\rangle\left|\psi_2\right\rangle+\left|\psi_2\right\rangle\left|\psi_1\right\rangle,$$

where $\left|\psi_1\right\rangle$ is packet from one source, and $\left|\psi_2\right\rangle$ is packet from another one.

This state says that the system is in a superposition of states, in each of which one of the particles comes from one source, and another necessarily from another source. I.e. the system is highly entangled. Such system could be created e.g. by some generator of pairs of particles with opposite momenta.

But two independent sources are clearly not such a source of entangled pairs. As the particles are indistinguishable, and there's no symmetry which would allow us to determine that the particles come from different sources, we can't say which source the particle has come from. If we watch single particle emitting from the sources, they might come one after another from different sources, or they could repeatedly come from single source, then several times from another one. I.e. there's no rule that if one particle is from source A, next detected one is from source B. So, the initial state must be in the following form:

$$\left|\Psi\right\rangle=\left(\left|\psi_1\right\rangle+e^{i\phi}\left|\psi_2\right\rangle\right)\otimes\left(e^{i\psi}\left|\psi_1\right\rangle+e^{i\chi}\left|\psi_2\right\rangle\right)+\\ +\left(e^{i\psi}\left|\psi_1\right\rangle+e^{i\chi}\left|\psi_2\right\rangle\right)\otimes\left(\left|\psi_1\right\rangle+e^{i\phi}\left|\psi_2\right\rangle\right),$$

where $\phi,\psi,\chi$ are constants, which depend on the experimental setup.

Now from the form of the initial state it's obvious that the interference pattern will be present, and it's confirmed by numerical simulation.

Thus, it appears that even in this multiparticle experiment particles interfere with themselves, rather than with each other, to produce visible pattern on the screen.

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If you generate two wave packets at $\pm \infty$ and direct them to each other, they will be constant in vertical planes, so you will not see interference. In a real experiment, with finite distance between the sources, you see that the time to get from the source (suppose it at the axis) is shorter than the time it takes to go a bit off axis. There are your interference rings.

The same applies if the sources are at infinity, but directed at a non-flat angle.

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  • $\begingroup$ You seem to have misunderstood my question. I've added some pictures to illustrate what I'm thinking of. If you understood it right, could you elaborate on how you generate the packets? $\endgroup$ – Ruslan Mar 18 '14 at 13:34
  • $\begingroup$ You are right, I misunderstood. Let me rethink it. $\endgroup$ – Davidmh Mar 18 '14 at 14:16
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Well, try looking at it this way. We get interference patterns in the classic 2-slit experiment because the single source used is phase-coherent with itself. If you then use a "trombone" delay line in the path to one of the slits, two things happen. First (doh), there's a phase delay which will cause the interference pattern to shift laterally. Second, if the delay is longer than the coherence length of the source, you'll lose the interference because the light in the two paths start having random phase differences.

Now, if you happen to have two independent sources (same wavelength) with extremely long coherence lengths, then their relative phase difference remains constant over your observation time, and you get interference patterns again.

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  • $\begingroup$ In classical 2-slit experiment we get interference pattern because each particle goes "through both slits", thus adding probability amplitudes of its wavefunction as a function in one-particle configuration space. But if we have two independent beams, no matter how long their coherence length, the particles in each move in their own configuration spaces, and don't interfere in the same way (and symmetrization doesn't help). Joint probability, of course, will have many minima and maxima, but the image on screen won't. Thus I don't understand how photons could interfere giving fringes on screen. $\endgroup$ – Ruslan Mar 18 '14 at 14:47
  • $\begingroup$ @Ruslan it's not always necessary to look at the problem as purely particle-based. Follow the wave formalism and you'll get the result which is experientally observable. $\endgroup$ – Carl Witthoft Mar 18 '14 at 15:57
  • $\begingroup$ Great, but the views must not contradict each other, right? I'm not trying to use classical particles or other irrelevant entities — just quantum mechanics for two bosons. $\endgroup$ – Ruslan Mar 18 '14 at 16:51
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I might be misinterpreting things, but I was under the impression that the notion of interference and coherence in light sources was typically interpreted in the context of classical statistical optics, rather than through quantum mechanical interference of individual bosons (although that may just be because I lack knowledge of quantum optics).

As an often-cited example, suppose light emitted from a quasimonochromatic source strikes $N$ pinholes at positions $\mathbf{q}_j$ for $1\leq j\leq N$ on a sheet, behind which is a viewing screen. Assuming that the field striking each pinhole is $V_j(t)$, the field emitted from pinhole $j$ is a spherical wave of the form $V(\mathbf{r},t)=K_jV_j(t)/|\mathbf{r}-\mathbf{q_j}|$ for some constant $K_j$. So at a point $\mathbf{p}$ on the sheet, the expectation value of the intensity is $$I(\mathbf{p})=\left<\overline{\left(\sum_{j=1}^N\frac{K_jV_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}{|\mathbf{p}-\mathbf{q}_j|}\right)}\left(\sum_{j=1}^N\frac{K_jV_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}{|\mathbf{p}-\mathbf{q}_j|}\right)\right>$$ $$=\sum_{j=1}^N\sum_{k=1}^N\frac{\overline{K_j}K_k}{|\mathbf{p}-\mathbf{q}_j||\mathbf{p}-\mathbf{q}_k|}\left<\overline{V_j\left(t-\frac{|\mathbf{p}-\mathbf{q}_j|}{c}\right)}V_k\left(t-\frac{|\mathbf{p}-\mathbf{q}_k|}{c}\right)\right>$$ $$=\sum_{j=1}^N\sum_{k=1}^N\frac{\overline{K_j}K_k}{|\mathbf{p}-\mathbf{q}_j||\mathbf{p}-\mathbf{q}_k|}\left<\overline{V_j\left(t\right)}V_k\left(t+\frac{|\mathbf{p}-\mathbf{q}_j|-|\mathbf{p}-\mathbf{q}_k|}{c}\right)\right>$$ $$=\mathbf{a}^\dagger\boldsymbol{\Gamma}\mathbf{a}$$ where $$\mathbf{a}_j=\frac{K_j}{|\mathbf{p}-\mathbf{q}_j|}$$ is the vector of field strength coefficients for the pinholes and $$\boldsymbol{\Gamma}_{jk}=\left<\overline{V_j\left(t\right)}V_k\left(t+\frac{|\mathbf{p}-\mathbf{q}_j|-|\mathbf{p}-\mathbf{q}_k|}{c}\right)\right>$$ is the correlation matrix element between the fields at $V_j$ and $V_k$ with time delay $\tau_{jk}=\frac{|\mathbf{p}-\mathbf{q}_j|-|\mathbf{p}-\mathbf{q}_k|}{c}$.

Note $\mathbf{a}^\dagger\boldsymbol{\Gamma}\mathbf{a}$ is a matrix quadratic form, and in the absence of classical correlation (in the time-averaged sense) $\boldsymbol{\Gamma}$ is a diagonal matrix, and thus the fields add together without interference in a positive semidefinite fashion. The classical notion of interference is really one of a statistical nature; when there are correlations between the EM field at different times, fields which travel different distances (and thus have different travel times) can add together on average in ways that wouldn't occur when there were no correlations between different times in the field.

Setting $N=2$ and interpreting $V_1$ and $V_2$ as the fields of the two lasers in question gives a simple expression for the case of pinhole interference between the beams. If the beams are "good enough" (ie have coherence times long enough to be measured, such as order of milliseconds) then $\boldsymbol{\Gamma}_{12}$ will be stable over the duration of milliseconds and the fringes will be observable.

As for how this coincides with a quantum description which you are looking for, I have no idea, but perhaps someone with more knowledge on the matter can weigh in. However, I think that the classical picture remains accurate except in the very low amplitude regime of single-photon-type experiments.

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