# What is the wavefunction of the localized particle?

1. Aug 12, 2009

### Jano L.

Hi everybody,
it is usually said that the wavefunction of the localized particle at position $$x_0$$ is

$$\psi_{x_0} = \delta (x-x_0).$$

Is there some good reason for this? I mean, this function cannot be normalized, which means that $$|\psi_{x_0}|^2$$ cannot be interpreted as a probability density. There is also problem that we can not use it to calculate the expectation value of position operator:

$$\langle \psi |x|\psi\rangle = \int \delta(x-x_0)x\delta(x-x_0) = ?$$

What if we tried to solve this problem by square-rooting the delta function? We would have

$$\langle \psi |x|\psi\rangle = \int \sqrt{\delta(x-x_0)}x\sqrt{\delta(x-x_0)} = x_0$$

What do you think? Why do we use delta function, which does not follow the rules for calculating expectation values?

2. Aug 12, 2009

### Demystifier

You are right that the square root of the delta function is a better representation of the position eigenstate.

But there is an even better formal way to deal with this stuff. Imagine that space is not a continuum, but a very dense lattice. You can take the lattice spacing as small as you wish, so it is a good picture of the continuum for all practical purposes. Now the position eigenstate is a Cronecker delta. It has a beautiful property that the square root of the Cronecker is equal to the Cronecker itself.

Alternativley, you may work directly in the continuum by working with approximate position eigenstates represented by Gaussians of a very small width. Their width can be taken as small as you wish.

3. Aug 12, 2009

### Jano L.

Thank you for the reply. Discrete space is a good idea - we get rid off the formal mathematical problems with normalization and infinities, but it is quite a strong physical assumption. What is the lattice and its spacing? This would be not Lorentz invariant... Even then, if we wanted to calculate the scalar product of two functions on the lattice, we would go back to the continuum model and integrate continuous functions. Although interesting, I think discrete model would be great change.

Let's stick to the continuum model for a while - then, my favourite for the localized wavefunction would be sharp normalized Gaussian, as you suggested. Then it is easy to calculate everything in the framework of QM and no problem arises.

But in textbooks, there is never $$\sqrt{\delta(x-x_0)}$$, but often $$\delta(x-x_0)$$. One of the arguments I heard is that delta is the eigenfunction of the position operator:
$$x\delta(x-x_0) = x_0\delta(x-x_0),$$
which is true in the distribution sense. If we had continuous normalizable wavefunction psi, the scalar product
$$\langle \delta_{x_0}|\psi \rangle$$
is defined, but

$$\langle \delta_{x_0}|\delta_{x_0} \rangle$$
is not defined. In contrary, for the square root of delta function, we have
$$\langle \sqrt{\delta_{x_0}}|\psi \rangle = 0$$

and

$$\langle \sqrt{\delta_{x_0}}|\sqrt{\delta_{x_0}} \rangle = 1$$

which leads to the conclusion that $$\sqrt{\delta(x-x_0)}$$ would be orthogonal to all normalizable wavefunctions, but itself, and therefore it is impossible to expand an ordinary function in these square roots.

Which assumption is more fundamental: normalizability (and therefore probabilistic interpretation of psi), or the fact that one should be able to expand ordinary wavefunction into a linear combination of eigenfunctions?

Or, are there other arguments that support delta against the square root of it?

4. Aug 12, 2009

### Demystifier

This is a good question. I would say the latter is more fundamental, but it depends on what exactly one wants to achieve. Anyway, with the latter approach there are no any problems in practice, because one never measures the particle position with perfect accuracy. In other words, position eigenstates are never descriptions of actual physical states.

5. Aug 12, 2009

### George Jones

Staff Emeritus
What kind of an animal (Where does it live?) is $\sqrt{\delta_{x_0}}$ that it can satisfy both
If $\sqrt{\delta_{x_0}}$ is either a state in a Hilbert space, or even a generalized state (functional on a Hilbert space), then the first condition (if true for all relevant $\psi$) gives that $\sqrt{\delta_{x_0}} = 0$.

6. Aug 12, 2009

### Jano L.

I was thinking about some sharp peak function like square root of normalized gaussian:

$$\psi_{x_0}(x) = \frac{1}{\sigma^{1/2}(2\pi)^{1/4} } e^{-\frac{(x-x_0)^2}{4\sigma^2}}.$$
This function has unity norm but its scalar product with well behaved function goes to 0 as $$\sigma$$ goes to zero. In the limit $$\sigma \rightarrow 0$$ it is maybe not an ordinary function from the usual function spaces, but this was the case also with $$\delta$$. Can we define this limit as the proper mathematical object?

Last edited: Aug 12, 2009
7. Aug 12, 2009

### George Jones

Staff Emeritus
Yes, the zero distribution.

8. May 9, 2010

### tpg

Consider dimensionally:

A delta function of x has dimensions of [L-1]. For correct normalisation, we require the spatial wavefunction to have dimension [L-1/2], suggesting that the square root is correct.

9. May 9, 2010

### tom.stoer

The problem you have with the delta function is due to the fact that you want to stay inside L². As soon as you allow for the closure of L² the problem disappears. Of course the wave function is no longer normalizable, but the formalism is still well definied.

Of course you can either deal with lattices, Gaussians etc. or you can try to enlarge L² appropriately.

Btw.: you have the same problem with plane wave states which are delta functions in momentum space.

10. May 9, 2010

### meopemuk

Both these properties are pretty natural. The latter one tells that probability of a particle localized at $$x_0$$ to be found at point $$x_0$$ is equal to 1, which is a tautology.

The former property means that the probability of a particle with a (continuous) wave function $$\psi$$ to be found exactly at point $$x_0$$ is zero. This is also true, because the "volume of the point'' $$x_0$$ is practically zero, and the chance of finding the particle there is negligible. However is we were able to sup up all these negligible chances over "all points in the space" we would get the total probability of 1.

When we use delta functions (instead of square roots of delta functions) for localized states, we are talking about "probability densities" at the point. These densities are generally non-zero and they can be integrated (instead of being summed over all points in space) to get the total probability of 1.

Eugene.

11. May 9, 2010

### LukeD

Actually... I don't think I've ever used either delta functions or square roots of delta functions directly.

I've only ever said: let's say we have a wave function |x> so that $$\hat x|x> = x|x>$$ and $$<x'|x> = \delta(x-x')$$

We can also take this as a limit of a "Gaussian" wave packet: $$\Psi(x) = \lim_{a \rightarrow 0} \frac{1}{\sqrt{\sqrt{\pi}a}}e^{-\frac{x^2}{2a^2}}$$
And this is exactly the square root of a delta function! The "Gaussian" wave packet is called such because $$|\Psi|^2$$ is a Gaussian, not $$\Psi$$

So the way I was taught does in fact use a $$\sqrt{\delta}$$

---

In one of my Quantum classes, we used a delta function potential well and talked about the solutions. To me this seemed strange because we ended up with only a single bound state that's not localized in the well (see http://en.wikipedia.org/wiki/Delta_potential). So it occurred to me a few days ago that we'd have to use $$\sqrt{\delta}$$ potential to get a bound state that's localized in the well. But I don't know what limit physically creates a square root of delta potential.