# Wave Function Collapse and Entropy

1. Jan 30, 2013

### stevendaryl

Staff Emeritus
I don't want to argue about whether the notion of "wave function collapse" is a good way of understanding quantum mechanics, or not. For the purposes of this discussion, let's just adopt uncritically the naive approach to quantum mechanics, that:

1. Between measurements, the system evolves according to Schrodinger's equation.
2. A measurement always produces an eigenvalue of the operator corresponding to the quantity being measured.
3. Immediately after a measurement, the wavefunction "collapses" to an eigenstate of that operator corresponding to the eigenvalue that you measured.

My question is: how is the entropy of a system affected by measurement? There is a sense in which it acts like a random perturbation, and so I would think that it would increase the entropy, but on the other hand, the state becomes more definite after a measurement, which would make me think that the entropy has been lowered.

Does my question make any sense, and if so, does it have a standard answer?

2. Jan 30, 2013

### atyy

http://en.wikipedia.org/wiki/Density_matrix#Entropy
"This entropy can increase but never decrease with a projective measurement, however generalised measurements can decrease entropy.[6][7] The entropy of a pure state is zero, while that of a proper mixture always greater than zero. Therefore a pure state may be converted into a mixture by a measurement, but a proper mixture can never be converted into a pure state. Thus the act of measurement induces a fundamental irreversible change on the density matrix; this is analogous to the "collapse" of the state vector, or wavefunction collapse."

Reference 6 of the Wikipedia quote is Nielsen and Chuang.
Theorem 11.9 (p515): Projective measurements increase entropy.
Exercise 11.15 (p515): Generalised measurements can decrease entropy.

Last edited: Jan 30, 2013
3. Jan 30, 2013

### Jano L.

It may help you to propose some example. What system? How exactly does the collapse happen? Which kind of entropy do you have in mind? Without that, I am afraid your question is too general.

4. Jan 30, 2013

Entropy is a statistical concept: S = k ln W (to quote Bolzmann) wherein the W irepresents the number of possible microstates corresponding to the macroscopic state of a system. For a single particle, W = 1, so S = 0.
I have always found it useful to recall that mathematically a single 'particle' is best represented by a tensor field in Minkowski space. That representation simultaneously satisfies both QM and GR.
When a 'measurement' is performed on a 'particle' or a 'wave', a tensor operator is applied to the tensor field to extract the desired value from the set of variables participating in the tensor field. Whether the value is appropriate for a 'particle' or a 'wave' depends on the effect the operator has on the tensor.
In any case, the Dirac mathematics are used: the tensor product of the conjugate tensor with the result of the application of the operator to the original tensor, which is then fully integrated over Minkowski space to produce the value sought. Whichever operator or field is used, the operation irreversibly alters the original tensor (in 3-D this is called 'wave collapse') into something that resembles either a standing wave or a moving particle.

5. Jan 30, 2013

### rkastner

I discuss the issue of 'collapse' in detail in my new book in terms of the Transactional Interpretation,

www.cambridge.org/9780521764155

In my proposal, wf 'collapse' is a form of spontaneous symmetry breaking. This would decrease the entropy relative to the perfectly symmetric state if you consider it analogous to the case, say, of a crystal forming in an amorphous solid.

6. Jan 30, 2013

### stevendaryl

Staff Emeritus
I read about the Transactional Interpretation decades ago in Analog magazine. Its inventor, John Cramer, was also a columnist in that magazine. I thought it sounded fascinating, but I haven't heard anything much about it since, so I assumed that it wasn't taken seriously by most physicists.

7. Jan 30, 2013

### rkastner

Hi Steven,

It is beginning to get more attention. I have been publishing papers on this; many of them are available on arxiv.org. Re the book, I have some introductory and preview material at my website, http://transactionalinterpretation.org/

Best regards
Ruth Kastner

8. Jan 30, 2013

### stevendaryl

Staff Emeritus
Ruth,

Thanks, I will take a look.

9. Jan 31, 2013

### Demystifier

This increase of entropy by projective measurement assumes that a true collapse does not exist.
But if it does, then it actually DECREASES entropy. See e.g.
http://arxiv.org/abs/1108.3080
TABLE 1

10. Jan 31, 2013

### kith

This thread is interesting, because opposite answers are given. I think given that stevendaryl explicitly asks about collapse, the answer is that entropy is the same after a measurement. When we perform a projective measurement on a pure state, the outcome is a pure state again and the (von Neumann) entropy of all pure states is zero.

Such a measurement can be devided into two parts: decoherence and collapse. The process of decoherence increases the entropy of the system and collapse reduces it. Nielsen and Chuang (cited by atty) only consider the first part, so there, entropy increases. But this doesn't produce a single outcome. Tegmark's "observation" in table 1 (cited by Demystifier) starts with the decohered state, so there, entropy decreases. But this doesn't cover the complete measurement process where we want to start with a pure state. If you combine the processes shown in table 1, my point that entropy is the same after the measurement is nicely illustrated.

It does become more definite only wrt to one basis, but becomes less definite wrt to other bases (consider spin measurements for example). So I don't think we should use this kind of reasoning.

11. Jan 31, 2013

### stevendaryl

Staff Emeritus
That's the basis for my original question. It seemed that if a particle was in a mixture of a spin-up state and a spin-down state, which is a high-entropy state, and then I measure the spin and find that it is spin-down, then afterward it's in a low-entropy state (pure spin-down).

12. Jan 31, 2013

### stevendaryl

Staff Emeritus
Okay, but in classical thermodynamics, the entropy of a system is roughly the log of the volume in phase space of the set of points consistent with macroscopic observables. Can this classical notion of entropy be affected by measurements?

13. Jan 31, 2013

### kith

That's right if you talk about an incoherent mixture of states (which corresponds to a diagonal density matrix). There, the situation is analogous to classical statistical mechanics.

But in this situation, there's no collapse. Collapse occurs only for superpositions, which are pure states and have zero entropy (analogous to points in phase space). If your initial state is a supersposition of a spin-up state and a spin-down state wrt to the z-axis, you can always find a measurement direction in such a way, that your state is an eigenstate to the corresponding spin operator. A measurement in this direction always yields the same result, hence your initial state already is a low entropy state. It's not meaningful to assign different entropies to superpositions and eigenstates, because the property of being an eigenstate depends on the basis. Every state is a superposition wrt to some basis.

Last edited: Jan 31, 2013
14. Feb 1, 2013

### Demystifier

All the discussion above seems to be about von Neumann entropy. The von Neumann entropy is zero for any pure state. Hence, if the initial and final states are pure states (whatever the unobserved intermediate states are), then the entropy of the final state is the same as that of the initial state.

Then why does entropy in Nature increase with time? Because the entropy responsible for that is NOT the von Neumann entropy. To get an entropy which increases with time, one must introduce some COARSE GRAINING. It is the corresponding coarse grained entropy (and not the von Neumann entropy) which increases with time.

15. Feb 1, 2013

### stevendaryl

Staff Emeritus

16. Feb 1, 2013

### atyy

Would this be the same as considering that the density matrix is a reduced density matrix (ie. we can only access the state of a subsystem)? So for a sequence of pure states of the universe, the von Neumann entropy of the reduced density matrix would still increase?

(I guess I'm asking if integrating out is a good enough form of coarse graining.)

Last edited: Feb 1, 2013
17. Feb 1, 2013

### Demystifier

If there was nothing resembling the wave function collapse, then one could say that entropy of the subsystem increases, due to decoherence. But the fact is that something resembling the wave function collapse does exist (which decoherence by itself cannot explain, which is why decoherence does not completely solve the measurement problem). For that matter it is not important whether the collapse is related to consciousness (von Neumann), or happens spontaneously (GRW), or is only an illusion (many worlds, Bohmian, etc.), as long as it exists at least in the FAPP sense.

18. Feb 1, 2013

### Demystifier

I think it's not.

19. Feb 1, 2013

### atyy

I guess it's not obvious to me that the reduced density matrix doesn't involve collapse, since I've seen it said that the Born rule is implicitly used in getting it.

20. Feb 1, 2013

### kith

But in most cases, there is nothing resembling wave function collapse. Decoherence occurs whenever the interaction between systems leads to entanglement and not only during measurements.

I'm still wondering how exactly this is related to the observable entropy increases. If we start with two pure states, I guess that any fundamental interaction that leads to maximal entanglement should begin to disentangle the systems afterwards. So I would expect an oscillating entropy for the systems (in classical mechanics, no entropy change arises from such a situation). Which of course would call for an explanation why our observations always take place in the rising entropy domain.

Last edited: Feb 1, 2013