# Generalized statistical interpretation

1. Apr 30, 2014

### lkijmj

I red griffiths many times but even now there is something I can't understand. It's about statistical interpretation. In his book chapter 3.4 he says

"If you measure an observable Q(x,p) on a particle in the state ψ(x,t), you are certain to get one of the eigenvalues of the hermitian operator Q(x,-ihd/dx)"

but when the particle is not in determinate state (I mean <σ^2>=0), we can't even get eigenvalue equation Qψ=qψ. So we don't know whether the observable is a eigenvalue of some eigenvalue equation or not.

Could you please explain the sentence above to me?

2. Apr 30, 2014

### Einj

Even if the initial state is not an eigenstate of Q, the result of the measure of Q is always going to be one of its eigenvalues. The probability of getting one of the possible eigenvalues will depend on the composition of your state (what mixture of eigenstates it contains).

Then, right after the measure, the state will collapse to the right eigenstate of Q (the one corresponding to the eigenvalue you just measures).

3. Apr 30, 2014

### grimx

Einj is right :)
I would also add that for principles of quantum mechanics, the eigenvalue of an operator is precisely the measure of that observable, then definitely after the measure will have an eigenvalue of the operator even if the state is not one of its eigenvector.

4. Apr 30, 2014

### WannabeNewton

A given observable $A$ will have a complete set of eigenstates $\{|a \rangle \}$ so that any state $|\psi \rangle$ is expandable in terms of this eigenbasis: $|\psi \rangle = \sum \langle a |\psi \rangle |a\rangle$ (where it is understood that if the spectrum is continuous, we replace the sum with an integral).

It is one of the postulates of QM that a "measurement" of $A$ will yield the eigenvalue $a$ of one of the eigenstates $|a \rangle$ with the probability (density if continuous variable) $|\langle a |\psi \rangle|^2$. This is a basic assumption of the theory that is elucidated at the very beginning of Griffiths and many other texts.

5. Apr 30, 2014

### Jilang

I think the issue is that when it's in a indeterminate state you won't get the same answer each time you measure it.

6. Apr 30, 2014

### Einj

Yes but that's not an issue. You can always repeat your experiment many times and determine the probability of obtaining a certain eigenvalue. This probability is completely determined by your initial state. There is no problem

7. Apr 30, 2014

### Jilang

But if were in a determinate state you would get the same result each time.

8. Apr 30, 2014

### Einj

Yes, but just because if you are in a determinate state this means that the probability of obtaining that particular eigenvalue is 1. It's all perfectly embedded in the quantum theory and it all perfectly predictable.

9. Apr 30, 2014

### strangerep

One sees this traditional fiction repeated so often that one becomes tired of correcting it. But... let's try again...

The notion that a state collapses to an eigenstate after a measurement is only plausible for "filter-type" measurements. For most other realistic types of measurements (i.e., nontrivial interaction between system and apparatus), it is plainly false. An extreme example is measuring the position of a photon by where it strikes a photosensitive screen. The photon is absorbed as part of the interaction with the screen, and does not even exist thereafter, hence is certainly not in an eigenstate of position.

For more extensive refutation of such bunkum-for-the-freshers, see Ballentine ch9.

[I hope I don't regret this.]

Last edited: Apr 30, 2014
10. Apr 30, 2014

### Useful nucleus

Thanks for this elaboration, strangerep! That is new to me.
However what you call "bunkum-for-the-freshers" is in many standard undergraduate texts for quantum mechanics. I believe the authors of these texts did not intend to be misleading, they simply tried to remain concise in a place where elaboration is needed.
What I still remember from the class lectures is that the professor completely refrained from stating or discussing the measurement postulate (This class was intended for nuclear engineers).

11. May 2, 2014

### Staff: Mentor

The issue here is books at the intermediate level tend to gloss over fine points. For example they will tell you a state is an element of a vector space. They aren't really. They are positive operators of unit trace. Only so called pure states can be mapped to a vector space.

THE book to get is Ballentine that has already been mentioned
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

It takes a lot of care to explain the statistical interpretation. Read it and it should be clear. For example QM can be developed from just two axioms and the collapse postulate is not one of them.

Thanks
Bill

Last edited by a moderator: May 6, 2017
12. May 2, 2014

### vanhees71

I've just looked at this Section 3.3 in Griffths's book. It's better than many other descriptions I know, but it's not optimal yet. It's crucial that the operators, describing observables are not only Hermitean but even (essentially) self-adjoint. This is nicely demonstrated in

F. Gieres. Mathematical surprises and Dirac's formalism in quantum mechanics. Rep. Prog. Phys., 63:1893, 2000.
http://arxiv.org/abs/quant-ph/9907069

Otherwise Griffiths seems not to mention the old-fashioned collapse postulate which is very problematic if not totally inconsistent with both the very foundations of causality (in the relativistic realm) and experimental practice in the real world of the lab. E.g., detecting photons with a CCD camera (or a good old photo scintillator or photographic film) usually leads to the absorption of the photon and not to its preparation as something localized at the position where the photon was detected. This view is particularly wrong for photons since it's not clear how to define a position observable for it at all, but that shouldn't bother you as a beginner in quantum theory.

The good point after all these debates about the state-collapse postulate, going on at least since the famous Einstein-Podolsky-Rosen paper and Bohr's response to it, is that it's not needed at all. The Minimal Statistical Interpretation is enough. Whatever you put on top of this minimal interpretation is metaphysics and subject to your private preference of a world view. The minimal interpretation is the core of the physics of quantum theory and not much disputed anymore, since more and more of the apparently "weird" features of quantum theory (mostly related with the phenomenon of entanglement, Bell's inequality, etc.) are confirmed to higher and higher accuracy.

As bhobba already recommended, a very good book is

L. E. Ballentine. Quantum Mechanics. World Scientific, Singapore, New Jersey, London, Hong Kong, 1998.

This you should, however, address only after you have a good grasp of how quantum theory is applied in physics. I don't know Griffiths's book so well, so that I cannot say, whether I'd recommend it. My favorite for the beginner's level is

J. J. Sakurai and S. Tuan. Modern Quantum Mechanics. Addison Wesley, 1993.

I think there is a more recent edition of this book, but I'm not aware what has been changed there.

Another great book is

S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press 2013.

There you find a nice discussion on the independence of the Born postulate (square of the wave function as probability distribution) from the other postulates of quantum theory.