Implication of eigenstates that are not normalizable

[TheCanadian]

If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable. I have been told that these eigenfunctions do not represent possible physical states, but what exactly does that mean? Is there a good interpretation of the physicality of these eigenstates?

If a measurement is done by an operator with either continuous or discrete spectrums, doesn't the wave function in either case still collapse to one of the eigenfunctions?

 

[Simon Bridge]

Look for an example of an hermitian operator with a continuous spectrum and examine it... see what it's properties are.

 

[bhobba]

If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable. I have been told that these eigenfunctions do not represent possible physical states, but what exactly does that mean?

They are idealisations introduced for mathematical convenience - they don't actually exist physically.

Its tied up with an advanced area of math called Rigged Hilbert Spaces:
https://en.wikipedia.org/wiki/Rigged_Hilbert_space

To work your way up to it I suggest the following:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Not just for QM but for any area of applied math its an essential bit of stuff in your tool-kit.

Once you have gone though that book how to apply it intuitively to QM should be straightforward, but if you have any problems post here.

Thanks
Bill

 

[vanhees71]

The "eigenvectors" for an "eigenvalue" in the continuum of the spectrum of a self-adjoint operator are never proper states, because these must be square integrable. This implies that you cannot ever prepare the system such that the observable represented by this operator take precisely this eigenvalue with certainty.

As an example take position and momentum for a particle moving in one dimension and use wavemechanics in the position representation. Then the Hilbert space vectors are represented by square-integrable wave functions $\psi(x)$, and the position operator is represented by multiplication with $x$ and the momentum operator by the derivative $\hat{p}=-\mathrm{i} \partial_x$ (setting $\hbar=1$ for simplicity).

Now we look for the eigenvectors of momentum we have to find a function $u_p(x)$ such that
$$\hat{p} u_p(x)=p u_p(x) \; \Rightarrow \; -\mathrm{i} \partial_x u_p(x)=p u_p(x) \; \Rightarrow \; u_p(x)=A_p \exp(\mathrm{i} p x).$$
Now for any $p \in \mathbb{R}$ this function is not square-integrable and thus not representing a state of the particle, but you can use it in the sense of a distribution to an appropiate subsystem of square-integrable functions ("test functions").

First let's determine the normalization factor $A_p$ in a convenient way by formally calculating the Hilbert-space product of two such generalized eigenfunctions:
$$\langle u_p|u_{p'} \rangle = \int_{\mathbb{R}} \mathrm{d} x A_p^* A_{p'} \exp[\mathrm{i}(p'-p)x]=(2 \pi) |A_{p}|^2 \delta(p-p').$$
Thus we choose $A_p=1/\sqrt{2 \pi}$.

Now take a square-integrable function $\psi(x)$, for which
$$\tilde{\psi}(p)=\langle u_p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x)$$
exists. This is the Fourier transform of $\psi$.

You can also easily show that
$$\langle \psi|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=\int_{\mathbb{R}} \mathrm{d} x |\tilde{\psi}(p)|^2,$$
i.e., the transformation from the wave function in $x$ representation to that in $p$ representation is unitary (at least on the subset of wave functions that is Fourier transformable).

You get the inverse transformation by
$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p u_p(x) \tilde{\psi}(p)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$
The physical meaning is that if you have a state represented by the wave function $\psi(x)$ in position space, the same state is represented by $\tilde{\psi}(p)$ in momentum space, and you now know how to transform from one to the other. This means that the probability distribution to find the particle at position $x$ is given by $|\psi(x)|^2$ and to have momentum $p$ is $|\tilde{\psi}(p)|^2$.

I won't go into the issue of collapse, because that would only trigger unnecessary discussions on interpretation...

 

[atyy]

The eigenstate is still good for using in the Born rule.

However, the collapse rule has to be generalized, since the eigenstate is not a legitimate state for the wave function to collapse to. In fact, even for discrete variables the collapse rule has to be generalized. One can see this from the fact that in Copenhagen, a measurement is subjective. Consequently, even if one uses the projection postulate, one can add an additional unitary operation after that as part of the "measurement".

The general collapse rule is found in http://arxiv.org/abs/0706.3526.

 

[bhobba]

If a measurement is done by an operator with either continuous or discrete spectrums, doesn't the wave function in either case still collapse to one of the eigenfunctions?

Of course. But the issue is it physically realizable or not. States with an exact position are not physically realizable but are introduced to get a mathematical handle on the situation. One can get by without it using the Von Neumann resolution of the identity formalism but its not as elegant.

Thanks
Bill

 

[ShayanJ]

This may be helpful.

 

[vanhees71]

The eigenstate is still good for using in the Born rule.

No! It's very important to stress that this is not true, because the probability distribution must be normalizable to 1 to make sense. E.g., the momentum eigenstate which in the position represtentation is given by
$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x)$$
is not square integrable and thus
$$|u_p(x)|^2=\frac{1}{2 \pi}$$
is not a probability distribution for $x \in \mathbb{R}$.

It would also violate the Heisenberg uncertainty relation between position and momentum components in the same direction,
$$\Delta x \Delta p \geq 1/2.$$
In fact you cannot even calculate the standard deviations of position and momentum inserting the plane wave instead of a proper square-integrable wave function (in the domain of $\hat{x}$ and $\hat{p}$), because the integrals are divergent.

A lot of confusion is caused by some other ideas, and this has nothing to do with interpretation but with mathematics (probability theory)!