# Does probability have units

1. Jul 9, 2009

### RedX

Whenever I see the amplitude for fermion scattering, it comes out dimensionless. However, for scalar scattering, the dimensions are 1/energy^2. For example, for 2 particles scattering into 2 particles in a $$\phi^3$$ interaction, there is just one internal line, so the amplitude is proportional to 1/(-mandelstam+m^2), where "mandelstam" is of the order energy^2. Is this right?

2. Jul 9, 2009

### Avodyne

In four dimensions, the phi^3 coupling has dimensions of energy, so the amplitude is also dimensionless.

In general, the dimension of the amplitude depends on the number of spacetime dimensions. Also, there are different conventions in use, especially for fermions, that can change the dimensionality of the amplitude.

3. Jul 9, 2009

### RedX

Ok, I see. In 6 dimensions, 2-2 scattering has dimensions 1/energy^2. But in 4 dimensions, it is dimensionless. This is true for all scalar interactions.

So if you have N particles, you integrate over $$d^Dx$$ for each one, giving a mass dimenension of -ND. However, in the LSZ formula there is a $$-\partial^2+m^2$$ for each particle, giving a mass dimension of +2N. Also for each particle you have a field $$\phi(x)$$ stuck in between the vacuum expectation value $$<0|T{\phi \phi \phi ...}|0>$$, and each field has dimensions (D-2)/2 to give +N(D-2)/2. Finally, the momentum conserving delta function that's taken out of the amplitude adds D to the total mass (since the delta function erases momentum when integrated, and we took it out, so this adds mass). The formula is then:

-ND+2N+N(D-2)/2+D=N+D-ND/2

So for 2x2 scattering:

4+D-4D/2=4-D

So in 4 dimensions the amplitude is massless, and in 6 dimensions the amplitude is -2.

What's interesting is that it not only depends on the dimension of space D, but also on the number of initial and final particles.

It's just really weird to think of probability or amplitudes as having dimensions, but o well.

Last edited: Jul 9, 2009
4. Jul 10, 2009

### Avodyne

It's because we're generally computing some sort of probability density rather than a probability. For example, in nonrelativistic QM, the wave function has dimensions of (length)^(-3/2) because it is $\int_V d^3x\,|\psi(x)|^2$ that gives the probability to find the particle in volume V.

5. Jul 10, 2009

### RedX

It doesn't seem to be this way however. Take spacetime to be 4-dimensional, so that there are 3 spatial dimensions, and take 2-2 scattering. The s-matrix amplitude, <f|i>, is dimensionless like you've said. The probability is |<f|i>|^2 divided by <f|f> and <i|i> (eqn 11.11 of Srednicki's book). <i|i> and <f|f> have mass dimensions of -4 (eqn 11.16 and 11.17 of Srednicki, plugging in V=-3). So all in all, the probability has mass dimensions 8.

This is a true probability, and not a density. You can integrate over all final states, but that does not introduce any mass dimensions (eqn. 11.19), as momentum space is peculiar in that you can have: sum over states=$$\frac{V}{(2\pi)^3}\int d^3k$$. The volume V cancels the mass dimensions of the integral d^3k.

In nonrelativistic QM, integrating in coordinate space seems to be different than momentum space in that in momentum space, you can multiply by the whole volume V to make the integral dimensionless if you want. However, in coordinate space, there is no whole momentum P to multiply d^Dx to make the integral dimensionless, so the wavefunction in coordinate space can't possibly be dimensionless, while the wavefunction in momentum space can.

edit: I just realized that if you include the delta function squared as in (11.12), it comes out to be dimensionless, cancelling the dimensions 8 said earlier. After careful consideration, it seems the probability, as defined in (11.11), is truly dimensionless, for any number of spacetime dimensions - you just have to take into account the mass change of the delta function when changing dimensions, both in the normalization of the states, and in the overall momentum conserving delta function.

Last edited: Jul 10, 2009
6. Jul 10, 2009

### diazona

It's my understanding that an actual probability must always be unitless, though probability densities can have various units depending on the formulas involved.

7. Jul 10, 2009

### Avodyne

|<f|i>|^2 divided by <f|f> and <i|i> is dimensionlees, but <f|i> is a d-dimensional delta function times the scattering amplitude, but not the scattering amplitude itself.

8. Jul 11, 2009

### RedX

Indeed.

How many dimensions a theory takes place in makes a big difference in the probability, even at tree level. So if the Feynman rules give the typical expression 1/(-mandelstam+m^2), then this is the same in all dimensions. However, the d-dimensional delta function, and the d-dimensional normalization that you divide by (<i|i> and <f|f>), changes the result so that the probability looks a lot different for the different dimensions.

9. Jul 12, 2009

### ibcnunabit

It may help to think of dimensions in terms of degrees of freedom.