Zafa Pi said:
However if the joint state is not a tensor product (i.e. the photons/state is entangled, or EPR). In that case neither individual photon has a state, so I don't see how you can talk about their amplitudes.
Alice and Bob don't have individual amplitudes, but the joint amplitude can be written as an amplitude-weighted sum of products of individual amplitudes. Let me explain the analogy with hidden variables for probabilities.
In terms of probabilities, we have a joint probability for Alice and Bob:
[itex]P(A,B|\alpha, \beta)[/itex]
where [itex]A[/itex] is Alice's measurement result and [itex]B[/itex] is Bob's measurement result, and [itex]\alpha[/itex] is Alice's detector setting, and [itex]\beta[/itex] is Bob's detector setting. A "hidden-variables" model for this joint probability would be a hidden variable [itex]\lambda[/itex] and probabilities [itex]P_{hv}(\lambda), P_A(A|\alpha, \lambda), P_B(B|\beta, \lambda)[/itex] such that:
[itex]P(A,B|\alpha, \beta) = \sum_\lambda P_{hv}(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)[/itex]
If there were such a hidden-variables model, then we could explain the joint probability distribution in terms of a weighted average (averaged over possible values of [itex]\lambda[/itex]) of products of single-particle probability functions. But alas, Bell proved that there was no such hidden-variables model for the joint probability distribution.
Now, let's shift the focus from probabilities to amplitudes. We let [itex]\psi(A, B|\alpha, \beta)[/itex] be the joint amplitude for the EPR experiment, where the amplitude is related to the probability via:
[itex]P(A, B|\alpha, \beta) = |\psi(A,B|\alpha, \beta)|^2[/itex]
So [itex]\psi(A,B|\alpha, \beta)[/itex] is a joint amplitude, but Alice and Bob do not have individual amplitudes. But is there a "hidden-variables" model for this joint amplitude? By analogy with the hidden-variables model for probabilities, we say that a hidden-variables model for the joint amplitude would be a hidden variable [itex]\lambda[/itex] and amplitude functions [itex]\psi_{hv}(\lambda)[/itex], [itex]\psi_A(A|\alpha, \lambda)[/itex], [itex]\psi_B(B|\beta, \lambda)[/itex] such that:
[itex]\psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda)[/itex]
If there were such a "hidden-variables" model for the probability amplitudes, we could interpret the joint amplitude as an amplitude-weighted sum of products of single-particle amplitudes.
It's not too hard to show that there is a hidden-variables model for amplitudes in EPR, even though there is no hidden-variable model for probabilities.
In the correlated two-photon EPR experiment, we have a joint probability distribution given by:
[itex]P(A, B|\alpha, \beta) = \frac{1}{2} cos^2(\beta - \alpha)[/itex] (if [itex]A = B[/itex])
[itex]= \frac{1}{2} sin^2(\beta - \alpha)[/itex] (if [itex]A \neq B[/itex])
where [itex]A[/itex] and [itex]B[/itex] are Alice's and Bob's measurement results, each of which have possible values from the set [itex]\{ H, V \}[/itex] (horizontal or vertically polarized, relative to the polarizing filter), and [itex]\alpha[/itex] and [itex]\beta[/itex] represent Alice's and Bob's filter orientations. In terms of amplitudes, we have:
[itex]\psi(A, B|\alpha, \beta) = \frac{1}{\sqrt{2}} cos(\beta - \alpha)[/itex] (if [itex]A=B[/itex])
[itex]= \frac{1}{\sqrt{2}} sin(\beta - \alpha)[/itex] (if [itex]A\neqB[/itex])
We can easily write this in the "hidden-variables" form [itex]\psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda)[/itex] by the following model:
- [itex]\lambda[/itex] has two possible values, [itex]0[/itex] or [itex]\frac{\pi}{2}[/itex].
- [itex]\psi_{hv}(0) = \psi_{hv}(\frac{\pi}{2}) = \frac{1}{\sqrt{2}}[/itex]
- [itex]\psi_A(A | \alpha, \lambda) = cos(\alpha - \lambda)[/itex] (if [itex]A=H[/itex])
- [itex]\psi_A(A | \alpha, \lambda) = sin(\alpha - \lambda)[/itex] (if [itex]A=V[/itex])
- [itex]\psi_B(B |\beta, \lambda) = cos(\beta - \lambda)[/itex] (if [itex]B=H[/itex])
- [itex]\psi_B(B |\beta, \lambda) = sin(\beta - \lambda)[/itex] (if [itex]B=V[/itex])
Using trigonometry, we can easily show that this satisfies the equation:
(In the case [itex]A=B=H[/itex]; the other cases are equally straight-forward)
[itex]\sum_\lambda \frac{1}{\sqrt{2}} cos(\alpha - \lambda) cos(\beta - \lambda)[/itex]
[itex]= \frac{1}{\sqrt{2}} cos(\alpha - 0) cos(\beta - 0) + \frac{1}{\sqrt{2}} cos(\alpha - \frac{\pi}{2}) cos(\beta - \frac{\pi}{2})[/itex]
[itex]= \frac{1}{\sqrt{2}} cos(\alpha) cos(\beta) + \frac{1}{\sqrt{2}} sin(\alpha ) sin(\beta)[/itex]
[itex]= \frac{1}{\sqrt{2}} cos(\beta - \alpha)[/itex]
So there is a strong sense in which amplitudes for quantum mechanics work the way we expect probabilities to work in classical probability.