# I Trying to model wave function collapse

1. Apr 11, 2017

### mike1000

Lets assume there is an observable represented by the operator, $\hat{A}$. It follow (I think) that the observed values and allowed states obeys an equation of the following form$$\hat{A}|\psi^i\rangle=\lambda_f|\psi_f\rangle$$ where $$|\psi^i\rangle = initial \ \ state \ \ vector$$ $$|\psi_f\rangle = final \ \ state \ \ vector$$My assumption is that the operator $\hat{A}$ operates on the state vector that exists immediately prior to the measurement.

Assume that the state vector, $|\psi^i\rangle$, is in a physical state that is some superposition of the eigenvectors, $\{ \psi_n\}$, of the operator $\hat{A}$
$$|\psi^i\rangle=\alpha_1|\psi_1\rangle + \alpha_2|\psi_2\rangle+\dots+\alpha_f|\psi_f\rangle+\dots+\alpha_n|\psi_n\rangle$$$$\hat{A}|\psi^i\rangle=\alpha_1\hat{A}|\psi_1\rangle+\alpha_2\hat{A}|\psi_2\rangle+\dots+\alpha_f\hat{A}|\psi_f\rangle+\dots+\alpha_n\hat{A}|\psi_n\rangle= \lambda_f|\psi_f\rangle$$$$=\alpha_1\lambda_1|\psi_1\rangle + \alpha_1\lambda_2|\psi_2\rangle+\dots+\alpha_f\lambda_f|\psi_f\rangle+\dots+\alpha_n\lambda_n|\psi_n\rangle = \lambda_f|\psi_f\rangle$$Which implies $\alpha_{i\neq f}= 0$ and $\alpha_f =1$

Which shows that the original assumption, that the initial state was a superposition of basis states, is false.(Proof by contradiction)

Last edited: Apr 11, 2017
2. Apr 11, 2017

### Staff: Mentor

That assumption is false in general. It will be true (to within a constant factor) only if $|\psi_i\rangle$ is an eigenvector of $\hat{A}$.

It's worth pointing out that your notation is an invitation to misunderstanding: You are using $|\psi_i\rangle$ to denote the initial state so $_i$ just means "initial", but later you are using subscripts inside the ket to denote the various eigenstates of $\hat{A}$ so $_i$ means "the i'th eigenstate of $\hat{A}$".

3. Apr 13, 2017

### Jilang

Sorry Mike, I cannot see why (1) would be true. What is the reasoning behind it?

4. Apr 13, 2017

### Physics Footnotes

@mike1000 In this and other posts you have made recently I see a recurring misunderstanding that, once cleared up, will probably answer all of your questions at once.

The problem is you are misunderstanding the fundamental notion and place of measurement in the quantum mechanical (QM) formalism. You sometimes speak, for example, of measuring the state vector or, as you do here, of a predicted result of a measurement as being a state vector or, as you have implied several times, as a measurement revealing the (pre-existing) state vector. In this post, you go even further and seem to treat the term $A\psi$ as representing a measurement process which physically 'transforms' the input state $\psi$ into some post-measurement state $\chi$. This interpretation of the formalism is completely wrong, and is leading you down all sorts of blind alleys. I'll explain why after a short sidenote, so you don't think I'm picking on you ;-)

SIDENOTE: This confusion is not your fault. Textbooks incorrectly overemphasize the post-measurement state as if it were part of the fundamental formalism. It isn't. In fact, most quantum experiments end in 'demolition measurements', meaning that the entity you started with (an electron or a photon, for example) is essentially destroyed (or transmuted or transformed) during the measurement process, never to be seen again. What is the state of a photon after its presence has been detected as a blotch on a photographic emulsion, for example? The question is ill-formed, as that photon has lost its identity.​

The QM apparatus is NOT designed to:
• predict the post-measurement state (this is an extra, but inessential, assumption people often like to throw into the cauldron),
• reveal the pre-measurement state (although with some cleverness you can infer this by taking sufficiently many judiciously chosen measurements), or
• model the measurement process itself (a problem which to this day has not been settled).
What does the basic QM apparatus do then?

As far as measurement is concerned, the QM formalism allows you to predict statistical distributions of outcomes (not states!) given an input state. The result of a measurement then is not a state but a value in some spectrum (typically, though not necessarily, the real numbers). In particular, for an observable $A$, the eigenvalue equation $A\psi=\lambda\psi$, should not be thought of as somehow determining a state, but rather as identifying a possible measurement outcome $\lambda$ for which you can later assign a probability $|\langle\psi|\phi\rangle|^2$ when a measurement is made on the system when it is prepared in the state $\phi$.

When you look at measurement from a value-outcome perspective, rather than a state-outcome perspective, I think you'll see why your issues with superpositions evaporate too. If not, I'm happy to have a go at convincing you further in this thread, but please make an effort to understand this post first.

Last edited: Apr 13, 2017
5. Apr 13, 2017

### Staff: Mentor

Just as a side-note to your side-note

The few observations that do not destroy what is being observed are called filtering type observations. They are much better understood as state preparation procedures than observations. These days state preparation procedures and states are pretty much synonymous - states are the equivalence classes of preparation procedures that give exactly he same outcomes from the Born Rule.

I rarely say this - but excellent post BTW.

As suggested by Gleason's theorem states are not fundamental - the observables are - states simply are an aid to calculating probabilities given an observable. They happen to form a vector space (at least the pure states do) so by fiat obey the principle of superposition - but that is just their mathematical structure.

Occasionally one makes use of the principle as in the link I often give on the double slit experiment, but mostly its just the mathematical nature of states.

BTW this is about the formalism of QM - interpretations have all sorts of views about the state - in some its very real (eg BM, MW),in others its the same as the formalism - just a mathematical aid (eg Quantum Bayesianism)

Thanks
Bill

Last edited: Apr 13, 2017
6. Apr 14, 2017

### vanhees71

One should make the really great posting #4 sticky somehow in the quantum-physics subforum and make it mandatory to read for any newcomer! I don't know, how often we discuss confusion, only because many textbooks don't make a clear distinction between states and observables and don't define these most important notions properly!

7. Apr 14, 2017

### ftr

Why stop there, go ahead be our guest.