# Should |<A|B>|^2 always be interpreted as probability?

1. Jun 18, 2013

In isolated time evolution, the Schrödinger equation takes one state ψ1 to ψ2. But it does this with certainty, no? So, if one uses the standard interpretation of the inner product as probability, would that say that <ψ2 | ψ1> = 1. But this would mean that the two states are equivalent, which they aren't. So, something is wrong here. I'd be grateful to find out what.

2. Jun 19, 2013

### lugita15

What you're missing is wave function collapse. According to (the standard interpretation of) quantum mechanics, the quantum state of a system keeps evolving deterministically according to the Schrodinger equation, but then when a measurement of the system is made, the quantum state spontaneously changes in a random way into some state with a definite value for whatever variable you measured. That's where probability comes in: if |ψ2> is a state with definite "zoinkness", and you measure the zoinkness of a system that was in state |ψ1> before measurement, then it will go collapse into a state with definite zoinkness, with a probability |<ψ2|ψ1>|^2 of going into the particular state |ψ2>.

3. Jun 19, 2013

Thanks, lugita15. Let me see if I have this straight: "probability" only has a meaning with respect to measurement; so if for example there is a closed undisturbed system, with ψ1 evolving to ψ2 evolving to ψ3 evolving to ψ4, and all the while no measurement is taken, one cannot use the concept "probability" between ψ2 and ψ3?

Last edited: Jun 19, 2013
4. Jun 19, 2013

### StarsRuler

The concept of probability can be used perfectly with no measurement, the probability is 1 that if in time 0 wavefunction is $\Psi_0$, then in time t, the wavefunction can be $e^{-iHt/hbar}\Psi_0$ of. If we introduce a measuring apparatus in the system, then we get the Born Rule by decoherence. The Born rule is always valid

5. Jun 19, 2013

Thanks, StarsRuler. That makes sense. Long live the Born Rule!

6. Jun 19, 2013

### vanhees71

What has been said in the previous posting, however, is not entirely correct. According to the fundamental rules of quantum theory, for a system prepared in a pure state, represented by the normalized Hilbert-space vector $|psi \rangle$, the probability to find values $(a_1,\ldots,a_n)$ for a complete set of compatible observables is given by
$$P(a_1,\ldots,a_n)=|\langle a_1,\ldots,a_n|\psi \rangle|^2,$$
where $|a_1,\ldots,a_n \rangle$ is a normalized common eigenvector of these observables, which is up to a phase uniquely defined, since the set of compatible observables are assumed to be complete.

It does not make sense to ask for the probability of system to be in another state $|\psi' \rangle$ when it is prepared in the state $|\psi \rangle$. It is important to distinguish between state kets and eigenvectors of observables!

7. Jun 19, 2013

### lugita15

Well, every state ket is an eigenvector of SOME observable, if only the outer product |ψ'><ψ'|.

8. Jun 19, 2013

### dextercioby

Yes, but it's not the point. Born's rule gives you the probability (density) only for a spectral value, not for a state.

9. Jun 19, 2013

### lugita15

Well, it gives both. It gives you the probability that the quantum state will collapse into a given eigenstate and spit out the associated eigenvalue.

10. Jun 19, 2013

### Jano L.

It depends, which Born rule we discuss.

For spins, the measurement gives definite value +1 or -1, and thus the spin state has to change either to z+ or z- .

For position of electron, the Born rule is different - the expression

$$|\psi(x)|^2$$

is probability density, not probability. The collapse postulate in the above form does not apply here. If we get $x_0$ by measurement, there is no $\psi_{x_0}$ function which would be eigenfunction to this value.

11. Jun 19, 2013

### lugita15

Yes, I was just simplifying things here.
Well, if you were able to make an ideal measurement of position, then you would indeed get a 100% precise value $x_0$, and the wave function would collapse into a delta function. But of course, in practice, you can't actually make an ideal measurement of position, you can only make a measurement of an observable like ∫|r><r|dV (or something of that general form) over a small region of space.

12. Jun 20, 2013

### StarsRuler

It makes perfect sense, except that a measurement requires a time that we would being despising. But in the original question, the born rule is to be applied after an evolution by the Schrödinger equation

Last edited: Jun 20, 2013
13. Jun 20, 2013

### vanhees71

Again, it does NOT make sense (within the usual quantum theory) not to distinguish between state kets and (generalized) eigenstates of self-adjoint operators that represent observables. The quantum theoretical state encodes probabilistic information of the outcome of measurements of observables in terms of probabilities (or probability distributions for values in the continuous spectrum of the operators) with the system given to be prepared in a state represented by the state ket (or more generally a statistical operator).

Formally you can see that it is indeed important to distinguish the two uses of bras and kets is the fact that the time evolution is only determined up to a time-dependent unitary transformation, which enables you to distribute the time dependence between state kets (statistical operators) and the self-adjoint operators representing observable quite arbitrarily. This gives rise to the different pictures for time evolution in use (Schrödinger picture: statistical operators carry the full time dependence, observable operators are time-independent; Heisenberg picture: statistical operators time-independent, full time dependence on observable operators; Dirac picture: arbitrary distribution of the time dependence among the statistical operator and the observable operators).

Of course, the physical quantities, i.e., the probabilities or probability distributions must be picture independent. Thus, to apply Born's rule to pure states, you always must take the (generalized) scalar product of a state ket with a (generalized) eigenvector of a complete set of compatible observables. If you want to know only the probability (distribution) for less than a complete set of observables, you just use such a complete set containing the observables of interest and sum (integrate) over the other observables in this complete set. In this way (and only in this way) the probabilistic description becomes unique, i.e., independent of the choice of the picture of time evolution, and this is also the way quantum theory is used in everyday lab life :-).

14. Jun 20, 2013

### dextercioby

You're mixing von Neumann's postulate with Born's rule. They are normally postulated separately, since there are formulations (interpretations) where Born's rule is used, but von Neumann's <state vector collapse> is ignored. One such interpretation is the one originated by Einstein and developed by Leslie Ballentine, the so-called <ensemble interpretation>.

15. Jun 20, 2013

### vanhees71

Yes, perhaps I should have mentioned this again, I'm a follower of the ensemble interpretation. I do not see what the collapse hypothesis is good for. To the contrary, in my opinion it only makes trouble. Of course, I am aware that the ensemble interpretation makes quantum theory an incomplete theory. Strictly speaking it's not applicable to cosmology, because the universe as a whole is a single object, and we cannot set up an ensemble of universes to make experiments on them. Then, within the ensemble interpretation, the idea of a "quantum state of the whole unierse" doesn't make sense. On the other hand, that's not new: We all know that one of the most fundamental puzzles in contemporary physics is, how to marry the most comprehensive space-time description (general relativity or some generalization of it) with the most comprehensive description of elementary particles and their interactions (relativistic quantum field theory) in a consistent way. In this sense physics is "incomplete", and thus also quantum theory as a (substantial) part of physics.

16. Jun 20, 2013

### Jano L.

I agree when it comes to positions of particles. But what about spin? It seems that collapse happens when atoms pass the Stern-Gerlach magnet, since one can verify that after measuring spin z of beam prepared in any state the beam remains in z+ or z-.

17. Jun 20, 2013

### vanhees71

The Stern-Gerlach experiment can very well be described using quantum theory. The preparation of a determined spin-z component is well understood using the quantum dynamics of the system. No collapse hypothesis is needed to understand this preparation procedure. For a nice paper on this, see

G. Potel et al, PRA 71, 052106 ͑2005͒

18. Jun 20, 2013

### Jano L.

Interesting, thanks.

19. Jun 20, 2013

### lugita15

vanhees71 and dextercioby, certainly there are a variety of interpretations of quantum mechanics, but I was assuming that since the OP was asking a basic question, they just wanted to know what the standard bare-bones formalism says.

20. Jun 20, 2013

### vanhees71

The minimal statistical interpretation is the standard bare-bone formalism. Anything on top is adding some personal belief on how to metaphysically understand the formalism. In my opinion that's an interesting question of its own, but it's not one of physics but of philosophy. To use quantum theory in physics you only need the minimal statistical interpretation.