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 &lt; x &lt; 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

 

[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.