About wavepacket unit and probability

[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.

 

[jtbell]

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 $$\int_{-\infty}^{+\infty} {|\psi|^2 dx} = 1$$

 

[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

 

[bhobba]

|ψ|² needs to have units of inverse meters

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

 

[cpsinkule]

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

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.

 

[bhobba]

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.

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

 

[cpsinkule]

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

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.

 

[bhobba]

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?

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

 

[cpsinkule]

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

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.

 

[bhobba]

mod psi squared is not a probability

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

 

[cpsinkule]

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

 

[bhobba]

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

That's not what he said.

Thanks
Bill

 

[stevendaryl]

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

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.

 

[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

 

[bhobba]

but for a continuous basis, $|x\rangle$ has dimensions of $L^{\frac{-d}{2}}$ so that $|x\rangle \langle x|\ dx^d$ is dimensionless.

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

Thanks
Bill

 

[stevendaryl]

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

Oh? Then I misunderstood what was being discussed.

 

[cpsinkule]

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

 

[bhobba]

Oh? Then I misunderstood what was being discussed.

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

 

[bhobba]

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

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

 

[cpsinkule]

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

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

 

[stevendaryl]

I think there's something that I'm not getting about this. Take a simple case of a particle confined to the finite region $0 < x < L$, then the lowest energy eigenstate is

$\psi(x) = A sin(\frac{\pi x}{L})$

You pick $A$ so that $\int_0^L |\psi|^2 dx = 1$. The answer is $A = \sqrt{\frac{2}{L}}$. So $\psi(x)$ is not dimensionless.

 

[cpsinkule]

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

Units of probability? ∫dx has units of length.

 

[bhobba]

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

|x> is a state represented by a ket. It has no units. <x'| is another state represented by a bra. It also has no units. <x'|x> is an inner product of states not a length.

The x is a label of the states - it does not change that states are dimensionless nor does it give their inner product dimensions.

Thanks
Bill

 

[cpsinkule]

|x> is a state represented by a ket. It has no units. <x'| is another state represented by a bra. It also has no units. <x'|x> is an inner product of states not a length.

The x is a label of the states - it does not change that states are dimensionless not does it give their inner product dimensions.

Thanks
Bill

Again, vectors do not have dimension, projections do. Radius vector on theta unit vector is an angle. Projection on x unit vector is a length. Same idea here.

 

[bhobba]

So $\psi(x)$ is not dimensionless.

That it is labelled by something with dimensions does not give it dimensions. The x here is simply a label.

If you don't agree simply see the integral formulation I gave.

Thanks
Bill

 

[bhobba]

Again, vectors do not have dimension, projections do. Radius vector on theta unit vector is an angle. Projection on x unit vector is a length. Same idea here.

Again I have claimed no such thing. BTW vectors can and sometimes do have dimension but that is irrelevant here - QM states are dimensionless.

Thanks
Bill

 

[PWiz]

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 $$\int_{-\infty}^{+\infty} {|\psi|^2 dx} = 1$$

So if we're talking about higher spatial dimensions, such as allowing the particle to move in 3 dimensions, then will the probability of finding the particle in a particular volume become $$P = \int \int \int_ V |\psi(r,\theta , \phi)|^2 dV$$ ? If so, then how will the normalization condition change? Will we integrate $\psi$ over an infinite volume?

 

[cpsinkule]

Again I have claimed no such thing. BTW vectors can and sometimes do have dimension but that is irrelevant here - QM states are dimensionless.

Thanks
Bill

Yes, a state is a vector in a vector space. The wave function is a glorified projection. Every value at each x is an expansion coefficient. Another example, the velocity vector has no units until you project its components out. Projected on x you get m/s. Projected on theta you get rads/s. An abstract vector has no units unless you project it onto a basis

 

[bhobba]

the velocity vector has no units until you project its components out.

That's incorrect. Velocity as a vector has dimensions distance/time.

Thanks
Bill

 

[cpsinkule]

That's incorrect. Velocity as a vector has dimensions distance/time.

Thanks
Bill

That's incorrect. It has units m/s in the Cartesian basis. In spherical coordinates you need a m/s and two angular velocities to specify it at any time

 

[bhobba]

That's incorrect. It has units m/s in the Cartesian basis. In spherical coordinates you need a m/s and two angular velocities to specify it at any time

You are wrong.

All this is doing is going around in circles. You are confused about some very basic stuff. But then again I suspect you will say I am.

There is no use going on so I will leave it there - maybe someone else can help you sort it out.

Thanks
Bill

 

[cpsinkule]

You are wrong.

All this is doing is going around in circles. You are confused about some very basic stuff. But then again I suspect you will say I am.

There is no use going on so I will leave it there - maybe someone else can help you sort it out.

Thanks
Bill

Actually, you're right the expansion coefficients I spherical coordinates all contain r to make them m/s, but the wavefunction has dimensions

 

[stevendaryl]

That it is labelled by something with dimensions does not give it dimensions. The x here is simply a label.

I didn't come to the conclusion that $\psi(x)$ has dimensions because it is labeled by $x$. I came to that conclusion because:

$\psi(x) = \sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})$

and $L$ has dimensions of length.

 

[stevendaryl]

I didn't come to the conclusion that $\psi(x)$ has dimensions of length because it is labeled by $x$. I came to the conclusion that $\psi(x)$ has dimensions $L^{\frac{-1}{2}}$ because:

$\psi(x) = \sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})$

and $L$ has dimensions of length.

 

[bhobba]

I'm sorry, but you are wrong. any textbook in the subject clearly states the dimensionaity of the wavefunction..

Really?

Exactly where does Ballentine say it? He examines the issue of probability in QM on page 55 if you want to look it up.

Thanks
Bill