Insights Why the Quantum | A Response to Wheeler's 1986 Paper - Comments

So the quantum weirdness in an EPR-type experiment is due to a combination of two things, neither of which is weird in itself:

1. Conservation laws (conservation of angular momentum)
2. Discreteness of measurement results (always getting ##\pm \frac{\hbar}{2}## for the spin measurement in any direction)

But it seems that there is something else going on in EPR, which is a collapse-like assumption: When you measure a fermion's spin along some axis ##\vec{a}##, then it is as if, afterward, it is definitely in that direction. That's different from an imagined classical measurement that is somehow constrained to give a discrete result. You could imagine (this is Bell's toy model) that the electron as an associated spin vector, ##\vec{s}##, and measuring spin with respect to an axis ##\vec{a}## would return ##+1/2## if the angle between ##\vec{s}## and ##\vec{a}## is less than 90^o, and ##-1/2## otherwise. This would give a discrete result, but the result would not be the actual spin vector of the electron.

 

I'm trying to get an intuitive understanding of the way that EPR probabilities (for anti-correlated spin-1/2 particles) are in some sense the closest we can get to the criteria:

1. The sum of the spins is zero.
2. Spin measurement always gives ##\pm 1/2##

If Alice measures her particle's spin along axis ##A## and Bob measures his particle's spin along axis ##B##, then it is impossible to satisfy both criteria, because unless ##A## and ##B## are aligned, none of the following combinations adds up to zero:

1. ##\frac{1}{2} (+\vec{A} + \vec{B})##
2. ##\frac{1}{2} (+\vec{A} - \vec{B})##
   
3. ##\frac{1}{2} (-\vec{A} + \vec{B})##
   
4. ##\frac{1}{2} (-\vec{A} - \vec{B})##

What the quantum probabilities do instead is the following:

• Filter only those events in which Alice gets +1/2. (That includes possibilities 1&2 above)
• Compute the vectorial average of the spin sums: This will be given by: ##\frac{1}{2} (P_1 (\vec{A} + \vec{B}) + P_2 (\vec{A} - \vec{B}))## (where ##P_1## is the probability of possibility 1 above, and ##P_2## is the probability of possibility 2).
• This average is still not zero, but its projection onto ##\vec{B}## is zero.

This uniquely determines the probabilities ##P_1## and ##P_2##:

1. ##P_1 + P_2 = 1##
2. ##(\frac{1}{2} (P_1 (\vec{A} + \vec{B}) + P_2 (\vec{A} - \vec{B}))) \cdot \vec{B} = 0##

The latter equation becomes:

##\frac{1}{2} (P_1 (cos(\theta) + 1) + P_2 (cos(\theta) - 1)) = 0## (where ##\theta## is the angle between ##A## and ##B##)

These equations have the unique solution: ##P_1 = \frac{1}{2} (1-cos(\theta)) = sin^2(\frac{\theta}{2})##, ##P_2 = \frac{1}{2} (1+cos(\theta)) = cos^2(\frac{\theta}{2})##

Those are the quantum probabilities for anti-correlated spin-1/2 particles.

That's sort of interesting, but my understanding of the motivation is a little muddled. I understand that you can't have perfect cancellation if the axes ##\vec{A}## and ##\vec{B}## are not aligned. But why ask for cancellation (on the average) along axis ##\vec{B}##?

 

I'm not sure, whether I understand your problem. This is an example for the fact that a single-particle quantity (like single-particle spin) in a many-body system can be determined for the system as a whole (here the total angular momentum) while the single-particle quantities are indetermined. That's an implication of entanglement.

In the here discussed case you have a total angular momentum 0 state of a two-particle system of spin-1/2 particles, i.e., the two-particle spin state is
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle - |-1/2,1/2 \rangle).$$
That's obviously a simultaneous eigenstate of ##|\hat{\vec{S}}^2 \rangle## and ##|\hat{S}_z \rangle## to the eigenvalues ##S=0##, ##\sigma_z=0##. Here
$$\hat{\vec{S}}=\hat{\vec{s}} \otimes \hat{1} + \hat{1} \otimes \hat{\vec{s}}.$$
Note that the ##S=0## state is very special, because in this case all three components of ##\vec{S}## are determined although these operators do not commute.

Nevertheless the single-particle spins are maximally undetermined, i.e., there probabilities are given by the Statistical operator
$$\hat{\rho}_A=\mathrm{Tr}_B |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1}$$
and
$$\hat{\rho}_B=\mathrm{Tr}_A |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1}.$$
Now measuring the angular momentum component at particles A and B in different directions you get the probabilities you quote, and that's all you know about the outcome of measurements of the single-particle angular momenta. Of course, the measured outcomes do not add up to 0. Why should they? Even in classical physics it doesn't make too much sense to add components of vectors with respect to basis vectors in different directions. Of course, if you measure the angular-momentum components for both particles wrt. the same direction, then they must add up to 0 due to the preparation of the two-body system in the ##S=0## state. As explained above here you have the special case of a preparation of all three angular-momentum components to have the determined value 0. This is special, because that's possible only for the ##S=0## state and is due to the complete rotational symmetry (isotrophy) of this one special state. So you can have sometimes common eigenvectors for incompatible observables, and that's the most common example for this fact.

I don't see any further specialty in this example, despite the fact that it's the most simple example to explain entanglement, Bell's inequality and all such unusual quantum correlations without a classical counterpart. It's only a problem, if you don't accept the quite abstract mathematical formulation of quantum theory and its minimal probabilistic interpretation in terms of Born's rule. Due to our persistent intuition from our experience with classically behaving (quantum) objects (aka many-many-many-...-body systems) we sometimes think we have to "explain" something more with quantum theory than there is contained in it, but that's pretty misleading.

 

The issue was not to derive the quantum probabilities from quantum mechanics, but to see if those probabilities can be derived from the assumptions that:

1. The measured angular momenta of the two particles separately yields a discrete answer: ##\pm \frac{\vec{A}}{2}## for the first measurement and ##\pm \frac{\vec{B}}{2}## for the second measurement.
2. The sum of the spins must add up to zero (in some average sense).

 

That's a different question, which is not describing the statistics you expect from quantum theory as detailed above!

 

Thanks for the warning. In general, philosophy on quantum theory rather confuses the reader than to help him or her. Maybe your book is an exception. Nevertheless the word philosophy in connection with quantum theory (or even physics in general) should be read as a caveat sign ;-)).

 

I'm not sure what you mean. Both the heuristic argument (not original with me; I was paraphrasing the Insights article) and the quantum theory make the same predictions: If Alice measures her particle's spin along axis ##\vec{A}## and Bob measures his particle's spin along axis ##\vec{B}##, then the conditional probabilities are:

• ##P_1## = the probability that Bob will measure spin-up given that Alice measures spin-up = ##sin^2(\frac{\theta}{2})##
• ##P_2## = the probability that Bob will measure spin-down given that Alice measures spin-up = ##cos^2(\frac{\theta}{2})##

(where ##\theta## is the angle between ##\vec{A}## and ##\vec{B}##).

 

Yes, but the spin components in non-collinear directions need not to cancel each other. Why should they?

 

That was what I asking RUTA for clarification about.

The sense in which there is cancellation on the average is this:

Among those events where Alice measures spin-up along direction ##\vec{A}##, the expectation for
##(\vec{S}_A + \vec{S}_B) \cdot \vec{B}## = 0

(where ##\vec{S}_A = + \frac{1}{2} \vec{A}## and ##\vec{S}_B = \pm \frac{1}{2} \vec{B}##, depending on whether Bob gets spin-up or spin-down)

 

It's again not clear to me what you want to calculate. The joint probability that A finds ##\sigma_A=\pm 1/2## and B find ##\sigma_B=\pm 1/2## is, of course
$$P(\sigma_A,\sigma_B) =|\langle \sigma_A | \otimes \langle \sigma_B |\Psi \rangle|^2.$$
Here ##|\sigma_A \rangle## and ##|\sigma_B \rangle## are the eigenvalues of the operators
$$\hat{\sigma}_{A}=\vec{A} \cdot \hat{\vec{\sigma}}, \quad \hat{\sigma}_{A}=\vec{A} \cdot \hat{\vec{\sigma}}.$$
I'm to lazy to explicitly figure this out, but I still don't see the point of the exercise :-(.