# About wavepacket unit and probability

1. Jul 19, 2015

### KFC

Hi all,
In quantum mechanics, we consider the squared modulo of a wave function has meaning of probability, so does it mean a wave packet should be unitless? I am reading some materials online about the Gaussian wave packet http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf, there at eq. 7.10, we have

$|\psi|^2 = \frac{1}{\sqrt{\pi}\sigma}\exp(-x^2/\sigma)$

I think $\sigma$ has the unit of meter so whatever inside the $\exp$ has no unit. But what about the coefficient? It has unit of 1/meter. When I am reading it, it looks like that $|\psi|^2$ is density of probability but not probability.

2. Jul 19, 2015

### Staff: Mentor

Indeed, $|\psi|^2$ is a probability density. In order to get an actual probability (which is unitless), you have to integrate over some interval in x: $$P = \int_a^b {|\psi|^2 dx}$$ gives the probability that the particle is in the range $a \le x \le b$, provided that $\psi$ is normalized so that $$\int_{-\infty}^{+\infty} {|\psi|^2 dx} = 1$$

3. Jul 20, 2015

### cpsinkule

|ψ|2 needs to have units of inverse meters because probability is dimensionless. remember that the dx in the integral has units of meters as well so
|ψ|2dx has no units

4. Jul 20, 2015

### bhobba

That's wrong.

Its dimensionless. Remember its the expansion of a state in position eigenvectors and since such expansions are arbitrary it must be dimensionless.

ie |u> = ∫|x><x|u> dx. By definition <x|u> is the wave-function and by construction obviously dimensionless as |x> and |u> are both a quantum state and have the same dimension. Quantum states are dimensionless - but that is just by the by.

Thanks
Bill

5. Jul 20, 2015

### cpsinkule

If you want to view it as a probability density then you would require it have units of inverse vol/area/length depending on the dimensionality of the problem.

6. Jul 20, 2015

### bhobba

The probability density comes from the Born rule which is independent of units.

It does not matter how you twist or turn |u> = ∫|x><x|u> dx. implies <x|u> has no units since |x> and |u> are the same thing. There is simply no way it can be anything else.

A probability density does not have the units of what its a density in - in fact its dimensionless.

Thanks
Bill

7. Jul 20, 2015

### cpsinkule

I disagree. Units depend on the basis you project onto. If you project a position vector onto the x unit vector you get a length. If you project it onto the azimuthal unit vector you will get an angle. Are you implying that integrating a dimensionless quantity over an interval of length will still leave you dimensionless?dx is not a dimensionless measure.

8. Jul 20, 2015

### bhobba

I am saying exactly what I said. Look at the equation |u> = ∫|x><x|u> dx. All you are doing in integrating is multiplying something by a number and summing. Now if you obtain the same thing as what you multiply and sum ie a state what you multiply must be dimensionless.

But if you want to look at it as a probability density its the same thing. For example the 1/2 probability you assign to the heads and tails of a one dollar coin does not give it the dimensions of dollars. Probabilities are, by definition from the Kolmogorov axioms, dimensionless. They are a number assigned to an event.

Thanks
Bill

9. Jul 20, 2015

### cpsinkule

mod psi squared is not a probability....for a given x, mod psi squared gives you a finite number, but I'm sure you know the actual probability of any given x is 0. You are correct, probabilities are dimensionless, but mod psi is NOT a probability. <x|u> has dimension inverse length, the measure dx has units length and returns the sum to dimensionless.

10. Jul 20, 2015

### bhobba

Ok - probability density - it changes nothing. Its still dimensionless.

What's the density of the Dirac measure? Its what's used to represent discreet distributions by densities.

Thanks
Bill

Last edited: Jul 20, 2015
11. Jul 20, 2015

### cpsinkule

It's not dimensionless. you liked jtbell's response which agrees with me. Clearly stated it's not dimensionless until you integrate it.

12. Jul 20, 2015

### bhobba

That's not what he said.

Thanks
Bill

13. Jul 20, 2015

### stevendaryl

Staff Emeritus
This is one of those cases where there is a distinction between $\sum_n |n\rangle \langle n|$ and $\int |x\rangle \langle x| dx$. For a discrete basis, $|n\rangle$ is dimensionless, but for a continuous basis, $|x\rangle$ has dimensions of $L^{\frac{-d}{2}}$ so that $|x\rangle \langle x|\ dx^d$ is dimensionless.

14. Jul 20, 2015

### bhobba

You have a function whose integral gives a probability. You have agreed probability is dimensionless. Differentiate a dimensionless quantity to give the original density and its still dimensionless.

Thanks
Bill

15. Jul 20, 2015

### bhobba

The dimensions of |x><x| is not the issue.

Thanks
Bill

16. Jul 20, 2015

### stevendaryl

Staff Emeritus
Oh? Then I misunderstood what was being discussed.

17. Jul 20, 2015

### cpsinkule

The units are in the integration measure and the function, not the basis vectors. ∫dx has units of length...

18. Jul 20, 2015

### bhobba

We are discussing the dimensions of wave-functions. By definition a wave function is <x|u> in the following:
|u> = ∫|x><x|u> dx

This is shorthand for the limit of a sum of |xi> where the |xi> goes to a continuum. Since that is exactly the same thing as |u> the weights in that sum must be dimensionless.

Thanks
Bill

19. Jul 20, 2015

### bhobba

The dx there has units of probability. Its a probability density and applies to events - not lengths. It is the EVENT you get an observation in the infinitesimal interval dx.

Thanks
Bill

20. Jul 20, 2015

### cpsinkule

<x'|x> is the delta distribution, delta has units inverse length