# Is probability a Quantum Observable ?

1. Jul 18, 2008

### dreynaud

Is probability a quantum observable ?

I'd like to obtain an observable operator (self-adjoint, hermitian) with correct eignevalues (the probability).

Have anyone heard of such an operator in QM ? And what about QFT ?
Why is there no such operator in classical textbook ?

2. Jul 18, 2008

### Crosson

The only response would be "Probability of what"?

3. Jul 18, 2008

### dreynaud

Good remark : so the operator should depends on something.

Say a probability of transition.
My point is that we effectively measure probabilities (of whatever is measurable, position, energy, momentum, transition...), so it should be an observable. no ?

4. Jul 18, 2008

### Crosson

It is possible to calculate the probability of a transition from one state to another, but this probability is not an observable --- it is something that you calculate rather then measure.

5. Jul 18, 2008

### dreynaud

Yes I know.

But the first success of the Quantum mechanics was to be able to predict the observed intensity of atoms' spetral rays whick is a direct chek of the probability of transitions. So probabilities are observed I think, not only calculated.

Could we say that probabilities are not quantum observable because they are (the only?) quantities that still share "classical" properties (they are well defined) : If it was a quantum observable, I guess there would be a complementary operator, uncertainty relation, etc...

Do you agree with that ?

6. Jul 18, 2008

### Crosson

No, it predicted the exact energy of the spectral lines, but the kind of time-dependent experiments that confirm our calculations of transition probabilities were not done until 50 years later.

The reason that 'probability' is not a quantum observable is that we never measure probability. Let me ask you, 'is probability a classical observable?'

7. Jul 18, 2008

### dreynaud

Ok we don't agree...

Means in the 1960's ? I've got an old book of Louis de Broglie (who was at first a science historian) published in 1941 that says exactly what I said : the main motivation to go beyound Bohr's model was the prediction of the intensities I mentioned. Bohr's model (which is not QM) predticed the right energy gaps, not the intensities. Sorry...

Yes there is no 'detector' of probability' if that's what you mean. It is not entering in the Hemiltonian etc...

However, considering the 2-slit experiment with a small intensity source, one count the impact somewhere, and is able to determine the probability. So it can be reached throught experiment. That's what I call 'measure or observe the probability'.

Anything looks wrong here for you ?

Last edited: Jul 18, 2008
8. Jul 18, 2008

### Crosson

The observable being measured in the double slit experiment is position. A large number of position observations allows you to determine a probability distribution for that position variable.

9. Jul 18, 2008

### dreynaud

Ok, we agree.

There is an observable (operator) for X and not for the probability (this is simply contemplation of the theory).

I'm perfectly ready to accept this, but I wanted to understand the reason that conduct to this.

It seems that there exists in QM 2 "kinds" of 'measurable' quantity (read obtained following an experimental method):
- the ones associated with an operator (the usual ones in QM and in classical theory; some of them have no classical counterpart);
- the probabilities which have no operator.

Furthermore :
- the 'No operator' for the probabilities is linked to the fact that there is nothing that permits a "one shot" measure of it as it is possible for usual quantum observables.

I think we agree on that conclusion ?

Note : Von Neumann suggested to describe measure process with the help of projector (on a position for example). The probability is then given as the mean value of it (similarly mean position is given by the mean value of the 'X' operator; so it suggested me to determine if it was really possible to find a "probability operator"; Neuman's projector seems too different of usual observables with the eigenvalue properties to be good candidates).

Regards.

10. Jul 18, 2008

### Fredrik

Staff Emeritus
Consider a system described by a 2-dimensional Hilbert space, and write an arbitrary state as $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. The probability that a measurement of the observable with eigenvectors |0> and |1> will yield the result corresponding to |0> is $P=|\alpha|^2$. Do you expect this Hilbert space to admit an observable with an eigenstate $|P\rangle$ for each $P\in[0,1]$? If there was such an observable, then the Hilbert space would be infinite-dimensional.

If you want to define a probability operator, you have to explain what that idea would mean first. I just did, for one particular kind of system, and it was easy to see that the idea doesn't work. Can you explain what a probability operator would mean for an arbitrary physical system? If you can, you can probably also prove that the idea doesn't make sense.

11. Jul 18, 2008

### dreynaud

Yes. But why not : consider the two states of a two slits experiment. It forms a 2 dimensionnal hilbert space; It can also be written as a vector in the infinite-dimensional space of the eigenvector of the position operator for ex.
(I hope I have understood your point).

Last edited: Jul 18, 2008
12. Jul 19, 2008

### Fredrik

Staff Emeritus
A Hilbert space is either infinite-dimensional or it isn't. The number of dimensions is defined as the maximum number of linearly independent vectors, so if that number is two, it can't also be infinite. I'm sure you agree with that.

In the case of the double slit experiment, the Hilbert space of the particle is infinite dimensional.

In the case of a non-relativistic spin-1/2 particle, the Hilbert space is also infinite-dimensional, but it can be split into two as if it's composed of two subsystems. One of the Hilbert spaces represents the spin and is 2-dimensional. The other represents everything else and is infinite-dimensional. What my argument shows is that reasonable assumptions about what a "probability observable" would be, imply that the 2-dimensional subspace is infinite-dimensional. That's a contradiction.

13. Jul 19, 2008

### dreynaud

Note however, that as you showed, a finite Hilbert space is not realist.
But whatever:
If one accepts your argument (the properies you mentioned for the operator P are reasonable, your argument is ok on mathematical gounds, so I think it should be !), it then shows that probability operator is mathematically inconsistent with actual theory.
From that, a physicist should then asks either the theory sould be modified, or the theory is sufficient. Previous posts permits to answer that theory is sufficient. That's why I prefer the previous arguments as a better explanation of the "no operator" for probabilities.

Thanks for having spent time on the question. That was a good remark.
Regards.