What does the state vector mean?

[mike1000]

The number of eigenvalues will not change. But for a given superposition in the unrotated frame, the rotated superposition may involve a different number of eigenstates and so perhaps a different number of "possible outcomes". (Fewer or more eigenstates may have vanishing probabilities.)

What you are referring to are not possible outcomes. There are only n possible outcomes. The coefficients (even if some are zero) of the basis are not defining a new possible outcome, they are defining a new state which shows us how to partition the probability amplitudes among the n-possible outcomes.

 

[mikeyork]

If you are in an eigenstate there is only one possible outcome in that frame. E.g. spin 1/2 in up state. Rotate the frame and you get two possible outcomes: up or down.

When you talk about possible outcomes you must specify the conditions under which you detect those outcomes. In QM that means specifying both the detection basis and the nature of the prepared state.

 

[mike1000]

If you are in an eigenstate there is only one possible outcome in that frame. E.g. spin i/2 in up state. Rotate the frame and you get two possible outcomes: up or down.

I'm sorry but you are confusing me again.

I would say that there were always two possible outcomes and your example corresponds to a state where the probability amplitude of being in one of the eigenstates was 1 and zero for all other possible outcomes.

 

[mikeyork]

I'm sorry but you are confusing me again.

I would say that there were always two possible outcomes and your example corresponds to a state where the probability amplitude of being in one of the eigenstates was 1 and zero for all other possible outcomes.

But you never made that explicit. I edited my last post by adding two sentences in anticipation of this response. Read it.

 

[vanhees71]

An aside for Vanhees71, can I refer you to 'The Principles of Quantum Mechanics' by PAM Dirac, bottom of page 71 in the 4th Ed, "we may look upon the representative of a ket |P> as a matrix with a single column by setting all the numbers <ξ1 -ξu | P> which form this representative one below the other." He makes a similar suggestion for the bra vector as a row on page 72. I know this text is a little old now so whether the misunderstanding has to do with some later edition that I do not possess.

Hm, finally you found a flaw in one of Dirac's writings :-(.

 

[stevendaryl]

Please, don't do that! I've seen such notation from time to time even in textbooks, but this thread shows, how confusing that is! It's of course not wrong, but it's confusing. My experience is that it is much better to keep to bra-ket notation strictly for the abstract vectors. It's the great advantage of the Dirac notation that it is representation independent, and you can do many general calculations in a much clearer way without introducing a basis/representation.

I'm not 100% in agreement. I believe that matrices helps enormously to understand the concepts of state vectors, operators, eigenvalues, etc. Then when you understand those concepts for concrete cases, I think it's easier to generalize to dealing with them abstractly. But that's a pedagogical question which I can't really answer.

This, perhaps somewhat overpedantic, notation helps a lot to avoid the kind of confusion we try to cure in this thread!

Well, I think there is a difference of opinion about this. Some of the confusion is cleared up by considering how the concepts would apply to simple concrete cases.

 

[vanhees71]

I'm not against using matrices (or even better wave functions) for concrete calculations and examples in learning QT. To the contrary, there's no other way to learn the concepts than to apply them to concrete cases, but I don't want to equal a Dirac ket $|\psi \rangle$ with a column (or a wave function), because this misses the important point, that the column or wave function consists of two parts, namely the state vector and the eigenvector of a self-adjoint operator, representing an observable. Given this as a orthonormal basis you have a one-to-one mapping between vectors and columns, but you shouldn't write an equality sign.

The same holds true for vectors of any kind. E.g., a position vector in classical mechanics is not a column of three numbers but it's one-to-one mapped to such a column with respect to a chosen basis.

 

[stevendaryl]

I'm not against using matrices (or even better wave functions) for concrete calculations and examples in learning QT. To the contrary, there's no other way to learn the concepts than to apply them to concrete cases, but I don't want to equal a Dirac ket $|\psi \rangle$ with a column (or a wave function)

I was careful to say that the column matrix was a "representation" of the ket, not that it was equal to the ket.

 

[victor94]

I recommend this paper: http://aapt.scitation.org/doi/pdf/10.1119/1.13586

Abstract:

A state vector is not a property of a physical system (nor of an ensemble of systems). It does not evolve continuously between measurements, nor suddenly ‘‘collapse’’ into a new state vector whenever a measurement is performed. Rather, a state vector represents a procedure for preparing or testing one or more physical systems. No ‘‘quantum paradoxes’’ ever appear in this interpretation. The formulation of dynamical laws may involve path integrals and/or S‐matrix theory.

 

[vanhees71]

I was careful to say that the column matrix was a "representation" of the ket, not that it was equal to the ket.

But you wrote an equation of the structure "ket=column". I just said that I'd avoid this notation, because it's confusing. Of course, formally it's not wrong, what you have written.

 

[stevendaryl]

But you wrote an equation of the structure "ket=column". I just said that I'd avoid this notation, because it's confusing. Of course, formally it's not wrong, what you have written.

Okay. So to be self-consistent, I should have used some relation symbol other than equality.

 

[vanhees71]

I think so, but as I said, some people might find this overpedantic :-).

 

[Adrian59]

I'm not sure whether the original thread reached a satisfactory conclusion. Having re-read the thread I do think the original problem was confusing the base vectors with outcomes. It depends on what you mean by outcome. My interpretation is the state vector does represent an quantum outcome. If the system is solely in an state of one of the base vectors, which is then an eigenstate, then the result of the system being in that eigenstate is the coefficient / eigenvalue for that base vector / eigenvector. As such there are n such outcomes as eigenstates. The situation is complicated when the final state is a superposition of two or more base vectors when the number of possible outcomes can be infinite. If by outcome you mean value of an observable then it is possible for more than one eigenvector to have a particular eigenvalue.

 

[vanhees71]

Sigh. Again: The state (as the name suggests) describes the preparation of the system, i.e., literally in which state it is in. Then you can do measurements on this system, and what the knowledge of the quantum states implies is the probability of outcomes of measurements. Also there is not "the basis" but infinitely many bases. The measured quantity selects which probabilities you want to calculate and thus which basis (namely, the eigenbasis of the self-adjoint operator representing the observable) has to be taken to calculate these probabilities.

There is also no complication in any choice of the basis. You can calculate the outcome of measurements, i.e., the probabilities with which values of the measured observable occur, in any basis you like. You only must express the state and the eigenstates of the observable-represebting operator in terms of the chosen basis.

 

[mike1000]

I'm not sure whether the original thread reached a satisfactory conclusion. Having re-read the thread I do think the original problem was confusing the base vectors with outcomes. It depends on what you mean by outcome. My interpretation is the state vector does represent an quantum outcome. If the system is solely in an state of one of the base vectors, which is then an eigenstate, then the result of the system being in that eigenstate is the coefficient / eigenvalue for that base vector / eigenvector. As such there are n such outcomes as eigenstates. The situation is complicated when the final state is a superposition of two or more base vectors when the number of possible outcomes can be infinite. If by outcome you mean value of an observable then it is possible for more than one eigenvector to have a particular eigenvalue.

The state vector does not represent a possible outcome. If it did there would always be an infinite number of possible outcomes. The state vector shows how the probability amplitudes are distributed among the n possible outcomes, where n is equal to the number of eigenvalues of the measurement operator.

The state vector is the probability amplitude distribution. It follows that there are an infinite number of probability distributions since there are an infinite number of state vectors.

 

[Adrian59]

Yes I can see your point maybe I should have said - my interpretation is the state vector does represent an quantum state. An outcome implies a result which may not be measureable say in two different polarizations, you only get a result from the axis of measurement chosen. Though mike1000 appears to have solved his problem - I think. So I'll leave this thread. I have picked up my 'Principles of QM' by PAMD in contributing to this thread which can't be a bad thing.