# Projection operator and measurement

1. Mar 2, 2015

### TrickyDicky

I'm aware there have been plenty of discussions about Copenhagen interpretation vs ensemble interpretations (myself I have always been more fond of the latter) but I intend to explore new perspectives and stick as much as possible to what QM practitioners do in practice as opposed to obscure metaphysical arguments.

It seems to me that when it is argued that the projection operator is always unitary(like in section 9.2 in Ballentine book for instance) unitarity is always(with one exception, that I explain below) assumed at the outset rather than derived by the physics and the mathematical analysis of experiments.

The exception refers to experiments that imply measurements with only two mutually orthogonal possible outcomes, best exemplified by the Stern-Gerlach experiment. But in this class of experiments we have a degenerate density matrix that is a multiple of the identity matrix, and there is only 2 possible different eigenvalues, no possible degeneracy of the discrete eigenvalues for the singlet state preparation, so the operator of the observable is also a multiple of the identity, and the projection is orthogonal, no "collapse" here.
This kind of experiments are very often used in textbooks as the main example of how QM works in general and it might lead to a limited view of the measurement projection operator when one is not dealing with that special case.

I understand why a non-unitary projection postulate might be too hard to swallow as it contradicts the postulates based on the Hilbert space but it appears that the solution of the ensemble interpretation amounts to just ignore individual measurements, something hard to justify in any empirical theory and even more in QM.

The postulates of a theory are supposed not be derivable but obtained from experiments, regardless if they are statistical or not. Otherwise one is restricting the validity of the theory to the point that some of the calculations and steps when applied to real problems are totally unjustified. In that sense the(nonunitary) projection postulate is necessary if one wants to give a complete set of postulates from experiments. Not including it would restrict QM to stationary states.
For example when doing time-dependent perturbation it is hard to dismiss the individual mesurement used to perform the perturbation as something just magically given, just "given the state".

Of course, a situation like this doesn't arise when analysing a Stern-Gerlach experiment, and it is very easy to convince oneself that there is no non-unitary projection using this experiment as an example. I'd say this is related to the special kind of measurement one performs when measuring spin restricted to one axis.

But for the more general type of measurement of observables that admit more than 2 eigenvalues, either discrete with degeneracy or continuous, there is a non-orthogonal projection involved, that can only be ignored if one simply assumes unitarity because the formalism says so while in practice getting results in QM often implies using individual measurements as starting point for calculations.
Not to mention the ensemble-only view leaves out entropy and irreversibility.

2. Mar 2, 2015

### Staff: Mentor

Projection operator is a concept not from QM but from linear algebra and by definition is unitary.

It's encoded in the first axiom of QM which says (and there are other equivalent forms) - A Von-Neumann measurement is described by a resolution of the identity Ei such that the probability of outcome i is determined by Ei, and only Ei. The Ei are of course, from the definition of a resolution of the identity, disjoint projection operators. If yi are the values associated with outcome i then one gets the Hermition operator O = ∑yi*Ei which by definition is called the observable associated with the observation.

Then one has the Born Rule that says a positive operator P exists, called the state of the system, such that the expected outcome of a Von Neumann observation is Trace(PO). Its easy to see the probability of outcome i is Trace(PEi).

The above can be generalised to so called generalised observations based not on resolutions of the identity but POVM's - that however is another story.

Thanks
Bill

3. Mar 3, 2015

### TrickyDicky

Was it not clear I was referring to non-unitary collapse as defined from the Copenhagen interpretation, dating back to one of Von Neumann 2 types of possible wave function change(unitary evolution and non-unitary measurement)? It is obvious such a change in the state doesn't follow strictly linear algebra. My point is that contrary to what it is usually held, that this is just a feature of CI, it would be necessary operationally for any QM calculation(with the exception I made about 2-dimensional Hilbert spaces operators and states) that has to explicitly handle individual measurement outcomes(I used the example of time-dependent perturbation problems).

4. Mar 3, 2015

### martinbn

There is some confusion here. Projection operators are not unitary (unless it is the identity operator). So all the speculations are empty.

5. Mar 3, 2015

### TrickyDicky

What confusion or speculation? Quotes would help.

Without qualification this sentence is either false or empty.

6. Mar 3, 2015

### martinbn

Well, the way I understood your post was that you think that projection operators are unitary (except in the two demensional case), and they are not, that's the confusion.

7. Mar 3, 2015

### Staff: Mentor

Drats - I forgot. That's correct.

Thanks
Bill

8. Mar 3, 2015

### Staff: Mentor

By definition a unitary operator preserves norm. The only projection operator that does that is the identity operator.

Thanks
Bill

9. Mar 3, 2015

### TrickyDicky

I wrote the opposite, I was referring to the collapse postulate, that is considered non-unitary, I said that the 2-dimensional case is an exception.

10. Mar 3, 2015

### kith

Just as a feedback: here are two other parts of your post which are pretty incomprehensible to me.

11. Mar 3, 2015

### TrickyDicky

12. Mar 3, 2015

### Staff: Mentor

It means they are disjoint ie the subspaces they project to don't overlap.

Thanks
Bill

13. Mar 3, 2015

### TrickyDicky

Thanks for the feedback, the firs paragraph is just sayin that the spin density matrix for a doublet is a múltiple of the identity(a matrix with diagonal :1/2,1/2) and one can always argue there
is no collapse in this case.

The second paragraph is just the Trivial fact that for actual QM problems(like the mentioned time-dependent perturbation of some interaction Hamiltonian) one is forced to use individual measurements in the calculations so it is hard to use the argument of an abstract ensemble to avoid dealing with individual measurements and thus with collapse.

14. Mar 3, 2015

### TrickyDicky

In other words there is no need to (re)normalize the state like it is in the higher dimensional case. The projection is orthogonal to begin with.

15. Mar 3, 2015

### kith

We really need to clarify some terminology. This thread has 14 posts and it consists almost entirely of misunderstandings between people.

What you write in your last post makes me think that you are using the term "orthogonal" as in "orthogonal matrix", i.e. as a property of an operator / linear map / matrix. This refers to vector spaces over the reals only. In QM, you need to talk about operators being unitary which is the corresponding concept for complex spaces.

Using the term "orthogonal" in QM in such a way is very confusing, because it makes people think of different things. In both real and complex spaces, the term "orthogonality" is used for a relative property between vectors (or subspaces): two vectors are orthogonal if and only if their inner product is zero. In this sense, one could say that two projection operators are orthogonal to each other if the subspaces they project onto are orthogonal to each other. This is what bhobba wrote in post #12.

So we need to talk about unitarity. Now, there's the issue raised by martinbn in post #2. Verify for yourself that no projection operator (except the identity) can be unitary. So Ballentine doesn't argue "that the projection operator is always unitary" in section 9.2. He doesn't even mention a projection operator there.

Last edited: Mar 3, 2015
16. Mar 3, 2015

### TrickyDicky

Ok, when I say orthogonal projection I mean a unitary projection.
Yes, that unitary projection is what I'm saying applies to the spin Stern-Gerlach case due to the fact that the density matrix of spin is a multiple(unphysical global phase factor) of the identity so it keeps its shape for any unitary transformation , and measurements give mutually orthogonal outcomes with 50% probability, can't this be described as a unitary projection(invertible) given the state is complex so it allows continuous changes of sign(±1/2) in the unit circle of the complex plane?

No , he doesn't, he tries to show that measurement is a unitary process. But many QM textbooks use (non-unitary)projection operators when discribing measurement, that's why I associated the concepts, I regret the confusion.

17. Mar 4, 2015

### TrickyDicky

I'll try and clarify what I mean about the action of the spin operator on a quantum state being equivalent to the action of the identity element and thus describable as a unitary projection. Since spin states before and after the action of the operator-diagonal matrix with eigenvalues(1,-1)- can be said to belong to the unitary circle group of all the complex numbers of absolute value one(unitary complex numbers), the operator acts as an identity element, being neutral for the states as members of the unitary complex numbers group, to wich spin states belong. And again the probabilistic outcomes(where the change of sign is not neutral) is described by the density operator wich is invariant(modulo a multiplicative factor) to unitary transformations(performed by the Pauli matrices).

18. Mar 4, 2015

### kith

Ok. Still, the terminology doesn't make much sense. The only unitary projection is the identity. Applying the identity to any state vector doesn't change anything, so "applying a unitary projection" is just a fancy way of saying that you do nothing.

Note that an operator which changes the phase isn't a projection operator in the first place (verify this for yourself).

What is the density matrix corresponding to the spin up state $|+_z\rangle$?

19. Mar 4, 2015

### TrickyDicky

See #17. The unitary projection "does nothing" to the quantum state.
It is just 1/2, the expectation value coincides with the probability density when there is only one nonzero entry in the diagonal of the density matrix, the significant thing here is that this is invariant to any unitary transformation in the spin case.

20. Mar 4, 2015

### kith

As I said: if it changes the phase, it isn't a projection. I also asked you to verify this for yourself. If you insist on using terminology in a way nobody else does, you won't be understood.

I gave you the feedback that what you write is hard to understand. I showed you examples where you use non-standard terminology and I asked you for some simple things. Instead, you responded by writing essentially the same as before. I am not interested in arguing with you, so I think I won't comment in this thread much longer.

21. Mar 4, 2015

### TrickyDicky

Is a change of phase for the spin state something physical?

22. Mar 4, 2015

### kith

No. Why are you asking this?

23. Mar 4, 2015

### TrickyDicky

Because my only point is that the spin operators act physically on spin states represented either by vector states or density operators leaving them invariant, and therefore equivalently to a unitary projection, I don't think that is so hard to understand. I am not obviously saying that a spin matrix is literally an identity matrix as you seem to interpret, just that physically(and that is what counts here no?) it acts like the identity.

24. Mar 4, 2015

### kith

Let me just point out the issues I have with this single sentence: First, there is no physical significance of "the spin operator acting on spin states". Neither the state preparation, nor the unitary time evolution, nor the measurement corresponds to this. Second, $\sigma_x$ for example flips $\sigma_z$-eigenstates, so it obviously doesn't leave them invariant up to a phase. Third, you are still using the wrong terminology of "unitary projections". Why do you do this? I told you that it is wrong. Why didn't you at least bother to look up the definition of a projection operator? This makes it really tiring to post here.

25. Mar 4, 2015

### TrickyDicky

It is the mathematical description of the map(the unitary transformation) from the initial state to the final state for any of the three processes you cite. I would say those processes are physical.

This involves a change of axis. It introduces another operator so I'm restricting the measurement to one axis

Fine, I admit my liberal use of this term is not appropriate. It was not meant to irritate, I can change it simply to "unitary measurement". Let's forget about projections, I was just using Copenhagen terminology for measurements by pure inertia.