# Physically realisable states and spectra

1. Apr 20, 2014

### cooev769

We've been assigned Griffiths QM for undergraduate physics. I don't particularly like it, but anyway.

It says that if the eigenvalues an observable are continuous then the eigenfunctions do not represent physically realisable states. So the eigenfunctions of the hamiltonian are discrete and therefore the stationary states represent real states. But for the position operator for example which must have the eigenfunction of the form g(x)=A dirac delta(x-y), this is continuous and hence is not a physically realisable state.

But I thought when you make a measurement on a wave function is collapses to exactly that the dirac delta function, which would indicate this does actually occur physically. Or does this just say that this is not a physically realisable wave function given it is not dependent on time and just sits there as at a point for all eternity?

Haha thanks.

2. Apr 20, 2014

### WannabeNewton

I couldn't understand most of what you tried to elucidate but the reason Griffiths says eigenstates of observables with continuous spectrum, for a given system, are not physically realizable is they might not be normalizable under the Hilbert space norm as is the case for the momentum eigenstates of a free particle for example. It's not really a problem in practice because we can either extend our space to a Rigged Hilbert space (http://arxiv.org/pdf/quant-ph/0502053v1.pdf) in order to accommodate these "generalized eigenfunctions" that are not normalizable under the Hilbert space norm, or we can just use box normalization (http://csma31.csm.jmu.edu/physics/giovanetti/quantum/Normmomentum.htm) and avoid the Dirac delta functions altogether.

3. Apr 20, 2014

### Matterwave

The delta-function distribution for x is certainly a idealization. Physically it is impossible to have 0 uncertainty in any measurement which can yield a continuity of results. If my observable O can have eigenvalues in a continuum, say [-1, 1] for simplicity, I can't physically choose out 1 (and exactly 1) number from that because a single number is of zero measure, and the probability of getting that number is 0 (almost 0? I'll let a mathematician figure that one out). We can only constrain the value to some range O=[o-do,o+do], so we will get a sum of eigenfunctions, rather than one eigenfunction itself.

4. Apr 21, 2014

### bhobba

The QM space isn't really a Hilbert space as some texts will tell you. It can be treated that way but most physicits, for practical reasons, use what is called a Rigged Hilbert Space. The physically reliazeable functions are all nice ie are continuously differentiable, vanish at infinity etc. For convenience this is extended to include linear functionals defined on such functions, and the Dirac delta function is such an object. It isn't really a function, but to any desired degree of accuracy can be approximated by a legit function with all the nice properties of the physically realisable functions. It's sort of like in calculus treating dx etc as very small actual numbers. You can go a long way doing that and just like in calculus you defer the correct treatment until later best to defer what's going on here.

When you feel the urge get some books on Didtribution theory to understand the proper treatment.

Thanks
Bill

Last edited: Apr 21, 2014
5. Apr 21, 2014

### cooev769

6. Apr 23, 2014

### Jano L.

Understandable, in light of the widespread teaching of the rule that says "upon measurement, the $\psi$ function instantaneously changes into eigenfunction of the operator that corresponds to the measurement". The coordinate operator $x$ does not have regular eigenfunctions, which is why you have stumbled upon this. The resolution is, the rule is not valid in the case of measurement of position. As above posters say, there is no such thing as measurement that will give you value of the coordinate with absolute accuracy - there is always some uncertainty. The mathematical aspect of this is also interesting: since there is no function $\psi$ that would obey $|\psi(x)|^2 = \delta(x-x_0)$, the theory of $\psi$ functions and the Born rule for $|\psi|^2$ is not capable to work with localized probability distributions.

7. Apr 23, 2014

### cooev769

Jano thank you very much. Made it very clear. So basically humans suck at measuring stuff.