# Units of wave function

1. Aug 13, 2012

### qsefthuko66

Do the units of a wave function vary? i have heard that it just joules. What do you think?

2. Aug 13, 2012

### qsefthuko66

i have also heard that the units are (length)-dN/2, where N is the number of particles in the system and d is the dimensionality of space. Is this true?

3. Aug 13, 2012

### The_Duck

The units of the wave function are whatever they need to be to make the norm of the wave function equal to 1 (and 1 is dimensionless).

4. Jun 28, 2013

### MonkeyDonkey

If $|\psi|^2$ is a probabilty and thus dimensionless, how can $\psi$ have units? Who have you "heard" this nonsense from?

5. Jun 28, 2013

### jeppetrost

Aha, but it's NOT a probability. It's a probability density, and thus indeed dimensionfull. But, as stated before, the dimension is simply "what it needs to be".

6. Jun 28, 2013

### Staff: Mentor

For a one-dimensional system, this is how you calculate a probability, which is unitless:

$$P = \int_a^b {|\psi|^2 dx}$$

Given that dx is a length, what units must ψ have?

Repeat for a three-dimensional system.

7. Jun 28, 2013

### Jazzdude

It's not entirely correct to say that the wavefunction requires certain units for the probability to work out. In fact, wavefunctions are quite agnostic about the units/dimensions used. This is because the state space is really a projective space, containing the C-rays and not C-vectors.

This is not obvious because we are used to the normalization convention, that is we make sure the norm of a state is equal to 1 whenever we construct or derive it. Unitary operators preserve that norm and it is generally quite simple to calculate probabilities from a given state.

However, this is just a convention, and if you really spell out the expression for probabilities in the projective space you rather get something like

$P = \frac{\int_S |\psi(r)|^2 dr}{\int_R |\psi(r)|^2 dr}$

where $S \subseteq R$. Any choice of units for $\psi$ can be seen to cancel in this fraction, just like any other factor. The denominator is only absorbed in the normalization convention inside the wavefunction. But the original state description in the projective space does not require the concept of dimensions and can therefore be chosen to be just dimensionless, independently of the dimension of the expansion parameter $r$.

Cheers,

Jazz

8. Jun 28, 2013

### Jolb

I think Jazzdude is right--nonnormalized vectors are allowed in the Hilbert space, so you are required to use his definition of P, and thus units don't matter.

One basic example that nobody mentioned is that you can actually pick the basis of Hilbert space with which to express your wavefunction. So instead of ψ(x), which according to some above posts should have units of 1/√length, you could equivalently express it in momentum space, so for a normalized momentum space wavefunction you'd have
1=∫|ψ(k)|2 dk, so its units would actually be √length in this basis [k here is wavenumber]. But this only follows from the normalization convention, which is optional, and Jazzdude's answer is the more rigorous one.

9. Jun 28, 2013

### Staff: Mentor

At the introductory level which I assume the OP is at unless he clarifies this, one usually assumes that wavefunctions are normalized when calculating probabilities. At that level, one doesn't usually talk about Hilbert spaces, projective spaces, etc., either.

At least, in the US.

10. Jun 28, 2013

### Ravi Mohan

Ok how do you, then, explain the overlap between two wavefunctions?(which should be dimensionless)
Edit:
Consider the equation
$$\int |x\rangle\langle x| dx = \mathbb{1}$$
Now $\varPsi (x)=\langle x|\Psi\rangle$

Last edited: Jun 28, 2013
11. Jun 28, 2013

### Jolb

I don't think there's any issue here. The dimensions would work out just fine, wouldn't they? Is your problem with Jazzdude's probability formula, rather than using the normalization convention? Let's look at the expression you've written down for the identity operator:
$$\int |x\rangle\langle x| dx = \mathbb{1}$$
I believe that this equation requires the basis kets |x> to be normalized. Here's a demonstration of that fact:

- - - - - - - - - - - - - - - - - -
Assume your equation for the identity operator is true:
$$\int |x\rangle\langle x| dx = \mathbb{1}$$
We can plug this equation into itself:
$$\mathbb{1}=\int \int |x\rangle\langle x|x'\rangle\langle x'| dx dx' = \int \int |x\rangle\langle x'| \left(\langle x|x'\rangle \right) dx dx'$$
Now recall the usual representation of position basis kets: [If this is confusing, see Griffiths Intro to QM, 2nd ed., p 105]
<x|x'> = δ(x-x')
This would reflect our usual convention of orthonormality. But if we relax normality and only preserve orthogonality, we have
<x|x'>=cδ(x-x')
for some complex number c. Plugging this into the big equation above yields:
$$\mathbb{1} = \int \int |x\rangle\langle x'| c\delta(x-x') dx dx' = c\int |x\rangle\langle x| dx$$
Hence c must be 1 for the identity operator equation to be true.
- - - - - - - - - - - - - - - - - -

Actually I think there might be a mistake in my proof since I was expecting c to be a unit complex number... can anyone help me to see if I have a problem here?

Last edited: Jun 28, 2013
12. Jun 28, 2013

### dextercioby

The units matter in Perturbative QFT. For a wave-function in 'ordinary' QM they are barely useful.