Projection operator and measurement

[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.

 

[bhobba]

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

 

[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).

 

[martinbn]

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

 

[TrickyDicky]

There is some confusion here... So all the speculations are empty.

What confusion or speculation? Quotes would help.

Projection operators are not unitary (unless it is the identity operator).

Without qualification this sentence is either false or empty.

 

[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.

 

[bhobba]

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

Drats - I forgot. That's correct.

Thanks
Bill

 

[bhobba]

What confusion or speculation?

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

Thanks
Bill

 

[TrickyDicky]

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.

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.

 

[kith]

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

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.

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".

 

[TrickyDicky]

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

Thanks
Bill

How about orthogonal projections?

 

[bhobba]

How about orthogonal projections?

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

Thanks
Bill

 

[TrickyDicky]

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

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.

 

[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.

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.

 

[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.

 

[TrickyDicky]

Ok, when I say orthogonal projection I mean a unitary projection.

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.

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?

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.

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.

 

[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 which spin states belong. And again the probabilistic outcomes(where the change of sign is not neutral) is described by the density operator which is invariant(modulo a multiplicative factor) to unitary transformations(performed by the Pauli matrices).

 

[kith]

Ok, when I say orthogonal projection I mean a unitary projection.

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).

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

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

 

[TrickyDicky]

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).

See #17. The unitary projection "does nothing" to the quantum state.

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

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.

 

[kith]

See #17. The unitary projection "does nothing" to the quantum state.

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.

 

[TrickyDicky]

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.

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

 

[kith]

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

No. Why are you asking this?

 

[TrickyDicky]

No. Why are you asking this?

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.

 

[kith]

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.

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.

 

[TrickyDicky]

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.

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.

Second, $\sigma_x$ for example flips $\sigma_z$-eigenstates, so it obviously doesn't leave them invariant up to a phase.

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

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.

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.

 

[Fredrik]

There's a theorem that says that if H is a Hilbert space and M is one of its (Hilbert) subspaces, then for all x in H, there's a unique pair (y,z) such that y is in M, z is in the orthogonal complement of M, and x=y+z. The orthogonal projection associated with the subspace M is the map $x\mapsto y$. If we denote this map by P, we can prove that P is linear and satisfies the conditions $P^2=P=P^*$.

Note that
$$\|x\|^2=\langle x,x\rangle=\langle y+z,y+z\rangle =\langle y,y\rangle+\langle y,z\rangle +\langle z,y\rangle +\langle z,z\rangle =\|y\|^2+\|z\|^2 =\|Px\|^2+\|z\|^2.$$ This implies that the norm of $\|Px\|$ is strictly smaller than the norm of $x$, unless $z=0$, i.e., unless $x\in M$.

A unitary operator on the other hand, is a bijective linear isometry. So if U is unitary, we have $\|Ux\|=\|x\|$ for all x in H. If an orthogonal projection P, for the subspace M, is unitary, then we must have M=H. But the orthogonal projection onto H is the identity map, as martinbn mentioned earlier.

Alternatively, you can define an orthogonal projection as a linear operator P such that $P^2=P=P^*$. Then you can show that a P that satisfies this alternative definition of "orthogonal projection" is an orthogonal projection also in the sense of the first definition. Specifically, it's the orthogonal projection associated with the subspace P(H).

 

[Fredrik]

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.

What do you mean by "unitary measurement"? The time evolution of a physical system isolated from its environment is unitary. A measurement breaks the isolation and causes the system to evolve in a non-unitary way. When the measurement is complete, the final state can typically be described as the result of a projection of the initial state onto an eigenspace of the self-adjoint operator that represents the measuring device.

 

[TrickyDicky]

The time evolution of a physical system isolated from its environment is unitary. A measurement breaks the isolation and causes the system to evolve in a non-unitary way. When the measurement is complete, the final state can typically be described as the result of a projection of the initial state onto an eigenspace of the self-adjoint operator that represents the measuring device.

Totally agree. This was the main point I wanted to make in the OP(It seems I did a lousy job). My claim was that this is not an interpretation-dependent feature of QM, but that if one stick to how QM is performed in practice, this non-unitarity is unavoidable when dealing with time-dependent quantum systems, a quantum measurement is by definition a time-dependent perturbation to a quantum system and everytime we calculate the response of the system to an individual measurement we are using that non-unitary process as starting point. By contrast in time-independent quantum stationary systems nothing observable ever happens, so if one restricts QM to these systems one can avoid the non-unitarity and use an interpretation such as ensemble's that ignores individual measurements. But for QM in general this cannot be done coherently when calculating in QM.

What do you mean by "unitary measurement"?

Now, this was the exception that originated so much confusion in this thread, because I was making a distinction between two-state systems and the rest. Probably the expression "unitary measurement" is totally misleading because what I meant is that in this case the measurement is different, it is not a perturbation of the system in the same way as commented above. I thought I could refer it to the generalization of measurement called POVM but I'm not really sure I fully understand POVMs.
So I guess what I mean is that they are not full measurements when restricted to an axis in the case of spin, and certainly the transformation from the state before the measurement(or the preparation) to the state after is unitary.

 

[kith]

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.

No. The change of state in preparations / measurements is given by the projection postulate (at least for practical purposes) and the change of state in time evolution is determined by the time evolution operator. None of these corresponds to the spin operator acting on the state.

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

Well, if you act with $\sigma_z$ on the (for convenience unnormalized) state $|+_z\rangle+|-_z\rangle$ you get $|+_z\rangle-|-_z\rangle$. These states are not related by a global phase either. But as I said: nothing in the SG-experiment corresponds to the spin operator acting on the state in the first place.

I think it is overdue that you show a reference which includes the maths of what you have in mind.

 

[TrickyDicky]

No. The change of state in preparations / measurements is given by the projection postulate (at least for practical purposes) and the change of state in time evolution is determined by the time evolution operator. None of these corresponds to the spin operator acting on the state.Well, if you act with $\sigma_z$ on the (for convenience unnormalized) state $|+_z\rangle+|-_z\rangle$ you get $|+_z\rangle-|-_z\rangle$. These states are not related by a global phase either. But as I said: nothing in the SG-experiment corresponds to the spin operator acting on the state in the first place.

I think it is overdue that you show a reference which includes the maths of what you have in mind.

You are right, and the reason is that in nrqm there is no spin, the standard Schrodinger equation doesn't predict spin, Pauli matrices are introduced "by hand", but one needs Dirac equation and spinors. What we can do with the SG experiment is prepare a state with a 100% probability of measuring say up spin, this measure is a purely unitary process.