About wavepacket unit and probability

[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

 

[stevendaryl]

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

I don't have Ballentine, but what I learned was the same as in Wikipedia:

The unit of measurement for ψ depends on the system, and can be found by dimensional analysis of the normalization condition for the system. For one particle in three dimensions, its units are [length]−3/2, because an integral of |ψ|2 over a region of three-dimensional space is a dimensionless probability.https://en.wikipedia.org/wiki/Wave_function#cite_note-8

https://en.wikipedia.org/wiki/Wave_function#cite_note-8

https://en.wikipedia.org/wiki/Wave_function

 

[bhobba]

and L has dimensions of length.

That doesn't give it dimensions of length - again its simply a factor introduced for normalisation and is a label ie in this case it is labelled by two things L and x.

Again I refer you to the definition of wave-function. From its definition it can't have dimensions of length. If it did its definition would make no sense.

Thanks
Bill

 

[bhobba]

I don't have Ballentine, but what I learned was the same as in Wikipedia:

Then I would have to say its wrong. By its DEFINITION it can't have units.

Added Later
I did a bit of a search and there seems to be some controversy about it some claiming it gains the units of the dimension of the space its used into make probability dimensionless. The trouble with that view is you run foul of its very definition.

Thanks
Bill

 

[Jazzdude]

The fact that the state space of QT is a projective space makes any non-zero factor on a state representation meaningless. That also holds for dimensional factors like units. Ergo the wavefunction is entirely agnostic of units, as they cancel when you apply the full form of the measurement postulate (and not the one for the pre-normalised wavefunction).

We've discussed this in depth in at least one other thread in the past.

Cheers,

Jazz

 

[vanhees71]

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

No! This is a very common mistake! The important point is that $|x \rangle$ is NOT a proper state but a distribution. By definition it's "normalized to a $\delta$ distribution, i.e.,
$$\langle x|x' \rangle=\delta(x-x').$$
This implies that the formal dimension of $|x \rangle$ is $1/\text{length}^{1/2}$.

The proper state $|\psi \rangle$ is normalized to 1,
$$\langle \psi|\psi \rangle=1.$$
Thus the proper dimension is 1. Together, the wave function is
$$\psi(x)=\langle x|\psi \rangle,$$
which has dimension $1/\text{length}^{1/2}$, as it must indeed be, because the normalization condition reads
$$1=\langle \psi|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle \psi|x \rangle \langle x|\psi=\int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \psi(x).$$
The $\mathrm{d} x$ has dimension $\text{length}^1$ and $|\psi(x)|^2$ the dimension $\text{length}^{-1}$, which makes the dimensions correct.

Note that this is true for any wave function wrt. generalized eigenvectors for operators with the eigenvalues in the continuous part of their spectrum. The corresponding wave functions squared are probability DENSITIES not probabilities!

 

[bhobba]

This implies that the formal dimension of $|x \rangle$ is $1/\text{length}^{1/2}$.

I can't follow that at all - x is simply a label |x> is a state and has no dimensions.

Thanks
Bill

 

[bhobba]

The fact that the state space of QT is a projective space makes any non-zero factor on a state representation meaningless. That also holds for dimensional factors like units. Ergo the wavefunction is entirely agnostic of units, as they cancel when you apply the full form of the measurement postulate (and not the one for the pre-normalised wavefunction).

When you posted that I didn't quite get it. But after thinking about it I think you are right. I now think a state is rather strange as far as units are concerned and can't be assigned units at all - its not that it has no units - it can't even be assigned any. Looking at |u> = ∫|x><x|u> dx you would think <x|u> had inverse units of length so <x|u> dx is dimensionless. But from |<x|u>|^2 dx being a probability you would think it had dimensions (1/length)^1/2 - weird.

Thanks
Bill

 

[stevendaryl]

That doesn't give it dimensions of length - again its simply a factor introduced for normalisation

Maybe we mean something different by "dimensions". To me, L is a length, for example: 5 meters. In the problem that I'm talking about, a particle is confined to a square-well, and L is the distance between the two ends of the well. It's not just a label, it's a length. The quantity \frac{2}{\sqrt{L}} should have dimensions length^{\frac{-1}{2}}. If I write \psi(x) = \frac{2}{\sqrt{L}} sin(\frac{\pi x}{L}), then that&#039;s a function that has dimension length^{\frac{-1}{2}}. Maybe there is something subtle going on, but the whole point of dimensional analysis is to give a simple sanity check about whether you&#039;re missing some conversion factor or something. If it&#039;s too subtle, then you&#039;ve defeated the purpose of dimensional analysis, it seems to me.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Again I refer you to the definition of wave-function. From its definition it can&#039;t have dimensions of length. </div> </div> </blockquote><br /> I didn&#039;t say length, I said length^{\frac{-1}{2}}.<br /> <br /> Isn&#039;t the probability interpretation of the wave function that the probability of finding a particle in the region between x and x+\delta x is given (approximately, for slowly varying wave functions) by:<br /> <br /> P = |\psi(x)|^2 \delta x<br /> <br /> So by dimensional analysis,<br /> <br /> [P] = [\psi]^2 [\delta x]<br /> <br /> where [...] means &quot;dimensions of&quot;. If \delta x has dimensions of length, and P is dimensionless, then \psi must have dimensions of length^-1/2

 

[vanhees71]

I can't follow that at all - x is simply a label |x> is a state and has no dimensions.

Thanks
Bill

Then I can't help. What's there to understand in a very simple dimensional analysis? Since by definition
$$\langle x|x' \rangle=\delta(x-x')$$
and the $\delta$ distribution always has the dimension of the inverse argument (i.e., 1/length in this case) the generalized ket $|x \rangle$ must have dimension $1/\text{length}^{1/2}$, so that the (formal) scalar product between two of them has the same dimension as the right-hand side.

As stressed in my previous posting, this leads to the correct dimension for the wave function, which must have dimension $\text{length}^{-1/2}$ either, so that its modulus squared has the correct dimension 1/length!

 

[bhobba]

Then I can't help.

You just did.

Now I get it.

Thanks
Bill

 

[jtbell]

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 $$P = \int \int \int_ V |\psi(r,\theta , \phi)|^2 dV$$ ?

Yes. (Or use e.g. $|\psi(x,y,z)|^2$ in Cartesian coordinates)

If so, then how will the normalization condition change? Will we integrate $\psi$ over an infinite volume?

Yes.

(sorry for the delay, I'm out of town traveling.)

 

[PWiz]

Yes. (Or use e.g. $|\psi(x,y,z)|^2$ in Cartesian coordinates)
Yes.

(sorry for the delay, I'm out of town traveling.)

Okay, thanks.

P.S. Happy traveling, and have a safe journey!

 

[KFC]

Hi all, I didn't realize that here is that many discussions going on here. It seems that some of your points make senses. My original thought is the $|\psi|^2$ is density of probability not probability, hence, it should be of unit 1/meter. If I am wrong, I wonder if the material I find here http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf is incorrect also. Because if you take $|\psi|^2$ and plug in the SI unit there, I got 1/meter.

 

[jtbell]

My original thought is the $|\psi|^2$ is density of probability not probability, hence, it should be of unit 1/meter.

Your thought is correct.

If I am wrong, I wonder if the material I find here http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf is incorrect also. Because if you take $|\psi|^2$ and plug in the SI unit there, I got 1/meter.

In your first post in this thread, you refer to equation 7.10 in that PDF. Immediately before that equation, it says "The probability density is:". What's the problem?

 

[KFC]

Your thought is correct.
In your first post in this thread, you refer to equation 7.10 in that PDF. Immediately before that equation, it says "The probability density is:". What's the problem?

Sorry that I didn't put my question clear. When I first learn QM in the text, I ALWAYS thought wave function is dimensionless for some reason I am always told my instructor that wave function is probability. So I keep that in mind for years. But recently I read several books myself for deeper review on the QM concepts. I have some thoughts about the wave function which should be interpreted as density of probability not probability itself. That's raise an another question is a wave function should carry unit or not. That's way I post the first thread here. I don't remember which book I read weeks ago but in which the author put a question that I rephrased as "Is it any significant to say that the probability of finding an electron at position $x_0$ is blablabla ... ?" He said no, we can only say that the probability to find an electron at $x_0-dx$ to $x_0+dx$ is estimated by $|\psi(x)|^2dx$. This statement leads me to think that the modulo square of wave function is density of probability so it should have unit of 1/meter if the position (or dx) is chosen in SI unit. Of course, we could make it unitless if we rescaled position to unitless. Frankly, I am not quite positive on that, I learn most of that myself from text so I want your guys to confirm my understanding.