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### small question in QM

Posted: **Thu Oct 01, 2009 7:50 am**

by **betelgeuse1**

I have a small question: when you talk about probability density in QM you call this abs(psi)^2. If you need the probability you multiply this with the volume element dV and if you whant the probability in a domain you integrate over that volume domain. If you whant to make your computations in spherical coordinates then your volume element is r^2*sin(theta)dr*d(theta)*d(phi) but the probability density still remains abs(psi)^2. If the function has spherical symmetry and it does not depend on theta and phi you can integrate out the sin(theta) and d(phi) and you get a 4*pi*r^2 out. Now, I think you call the probability density also abs(psi)^2 and 4*pi*r^2*dr is the volume element. My problem: There was a small confusion about this, somebody told me that the probability density was abs(psi)^2 *4*pi*r^2, and I still don't agree with that. I mean if you write in electromagnetism Integral(rho * dV)=Integral(rho * 4*pi*r^2 dr) then the charge density is still rho, or not? (I mean of course if rho depends only on r). I just whant to make sure the conventions are the same... would be silly to make mistakes like this in the GRE...

### Re: small question in QM

Posted: **Thu Oct 01, 2009 9:06 am**

by **blackcat007**

betelgeuse1 wrote:I have a small question: when you talk about probability density in QM you call this abs(psi)^2. If you need the probability you multiply this with the volume element dV and if you whant the probability in a domain you integrate over that volume domain. If you whant to make your computations in spherical coordinates then your volume element is r^2*sin(theta)dr*d(theta)*d(phi) but the probability density still remains abs(psi)^2. If the function has spherical symmetry and it does not depend on theta and phi you can integrate out the sin(theta) and d(phi) and you get a 4*pi*r^2 out. Now, I think you call the probability density also abs(psi)^2 and 4*pi*r^2*dr is the volume element. My problem: There was a small confusion about this, somebody told me that the probability density was abs(psi)^2 *4*pi*r^2, and I still don't agree with that. I mean if you write in electromagnetism Integral(rho * dV)=Integral(rho * 4*pi*r^2 dr) then the charge density is still rho, or not? (I mean of course if rho depends only on r). I just whant to make sure the conventions are the same... would be silly to make mistakes like this in the GRE...

Probability density is abs(psi)^2

(abs(psi)^2)*d^

r is the probability of finding the particle between

r and

r+d

rintegrate this from a to b to get the probability between a and b

, for spherical symmetry probability between a and b is integral(abs(psi)^2 *4*pi*r^2)dr

abs(psi)^2 *4*pi*r^2 is just the term that we get after integrating twice for theta and phi, for spherical symmetric functions. I don't think there is any name to this term.

as for your second question, yes charge density is still rho, it is not rho * 4*pi*r^2

### Re: small question in QM

Posted: **Thu Oct 01, 2009 9:16 am**

by **betelgeuse1**

thanks, it's how I thought as well...

### Re: small question in QM

Posted: **Thu Oct 01, 2009 9:26 am**

by **physics_auth**

betelgeuse1 wrote:thanks, it's how I thought as well...

You probably refer to me... as i told you this is the RADIAL probability density. By multiplying this with dr you find the probability of finding the particle between r and r+dr independently of θ and φ. This independence is achieved by integrating in spherical coordinates over all values of φ and θ which leads to the multiplicative factor of 4π. See also my answer to Blackcat007 ... .

The probability density (not the radial one) is always Ψ*Ψ as you said ...

### Re: small question in QM

Posted: **Thu Oct 01, 2009 9:33 am**

by **physics_auth**

blackcat007 wrote:Probability density is abs(psi)^2

(abs(psi)^2)*d^r is the probability of finding the particle between r and r+dr

integrate this from a to b to get the probability between a and b

, for spherical symmetry probability between a and b is integral(abs(psi)^2 *4*pi*r^2)dr

abs(psi)^2 *4*pi*r^2 is just the term that we get after integrating twice for theta and phi, for spherical symmetric functions. I don't think there is any name to this term.

as for your second question, yes charge density is still rho, it is not rho * 4*pi*r^2

The radial probability density (i.e. of finding the particle at a radial distance r independently of the angular direction in the three dimensional space) is always 4π*Ψ*Ψ r^2, where it is assumed that Ψ has spherical symmetry, i.e. Ψ=Ψ(r) only.

If you take 4π*Ψ*Ψ r^2 dr you have the RADIAL probability (not the density) of finding the particle within a concentric spherical shell of radial width equal to dr and with internal and external radii r and r+dr respectively.

If you integrate from r_a to r_b > r_a you find the probability that the particle is in a spherical shell (centered say at O) with internal radius r_a and external radius r_b > r_a. Radial probability density has meaning when there is spherical symmetry ... .

### Re: small question in QM

Posted: **Thu Oct 01, 2009 12:24 pm**

by **blackcat007**

physics_auth wrote:The radial probability density (i.e. of finding the particle at a radial distance r independently of the angular direction in the three dimensional space) is always 4π*Ψ*Ψ r^2, where it is assumed that Ψ has spherical symmetry, i.e. Ψ=Ψ(r) only.

If you take 4π*Ψ*Ψ r^2 dr you have the RADIAL probability (not the density) of finding the particle within a concentric spherical shell of radial width equal to dr and with internal and external radii r and r+dr respectively.

If you integrate from r_a to r_b > r_a you find the probability that the particle is in a spherical shell (centered say at O) with internal radius r_a and external radius r_b > r_a. Radial probability density has meaning when there is spherical symmetry ... .

but in 3 dimension Integral{a->b}(Ψ*Ψ d^3

r) is the probability of finding the particle between a and b

and by definition Ψ*Ψ is the probability density, is the word radial that is making the difference?

like the radial probability density is 4π*Ψ*Ψ r^2 for spherical symmetric case . in that case will this term lose its meaning if i used say cylindrical coordinates?

### physics vs. grammar

Posted: **Thu Oct 01, 2009 1:35 pm**

by **betelgeuse1**

As far as I understand if I have cylindrical coordinates then 4*pi*r^2 has no meaning as the symmetry changed. It's not anymore a spherical symmetry but a cylindrical one. I used this facts but I never called them "radial probability density"... clear now!

I was also amused by the difference made in US lectures between "impulse" and "momentum"... I know now that momentum is p and impulse is what we call Delta(p) or F*delta(t) F=const. what we usually call change of impulse... But this is only grammar... nothing essential... still can make some difference in the GRE...

### Re: small question in QM

Posted: **Thu Oct 01, 2009 3:06 pm**

by **physics_auth**

blackcat007 wrote: but in 3 dimension Integral{a->b}(Ψ*Ψ d^3r) is the probability of finding the particle between a and b where a = (x1, y1, z1) and b = (x2, y2, z2)

and by definition Ψ*Ψ is the probability density, is the word radial that is making the difference?

you probably mean d^3r = dV = dxdydz = {in spherical coordinates} = r^2 sinθ drdθdφ where

r = radial coordinate, θ = polar coordinate, φ = azimuthal coordinate -> that is how they are defined in maths.

the term "radial" refers to coordinate r in spherical coordinates. When you are asked for the radial probability density this means that they ask you about 4π * r^2 Ψ*(r)Ψ(r) = ρ(r) = radial probability density.

like the radial probability density is 4π*Ψ*Ψ r^2 for spherical symmetric case . in that case will this term lose its meaning if i used say cylindrical coordinates?

In cylindrical coordinates, when some people refer to the "radial" coordinate, then they mean the distance from the axis of symmetry (not from a single point as in spherical coordinates). In this case the volume element dxdydz -> becomes -> ρ dρ dφ dz where ρ is the dinstance I said previously, φ = the azimuthal angle and z = the known z coordinate. Because ρ is different from r (used in spherical coordinates) the term "radial" is inappropriate (or if mentioned it refers to the distance from the symmetry axis -> ρ). However, cylindrical symmetry is not encountered in the stuff for the PGRE, it is commonly met in scattering theory which is NOT included in the material that one has to read for the test ... .

### Re: small question in QM

Posted: **Fri Oct 02, 2009 1:24 pm**

by **betelgeuse1**

You can speak also about bi-elliptical coordinates the basic fact is that you integrate over all the variables you are not interested in. In spherical, if you are interested about the value of your quantity considering only its "radial" coordinate you "average" out the other variables. After integrating over them they become constant parameters and you got it. The fact is easy and quite used in stat-mech. I was somewhat confused about the term used in the problem "radial probability density"... probably because I did not read the term "radial"... AM I AN IDIOT!!!

### Re: small question in QM

Posted: **Fri Oct 02, 2009 2:06 pm**

by **physics_auth**

betelgeuse1 wrote:You can speak also about bi-elliptical coordinates the basic fact is that you integrate over all the variables you are not interested in. In spherical, if you are interested about the value of your quantity considering only its "radial" coordinate you "average" out the other variables. After integrating over them they become constant parameters and you got it. The fact is easy and quite used in stat-mech. I was somewhat confused about the term used in the problem "radial probability density"... probably because I did not read the term "radial"... AM I AN IDIOT!!!

No you are simply careless ... . Be careful in the official test. To make errors while practising at home doesn't harm as long as you don't repeat them in the real test date!

### Re: small question in QM

Posted: **Fri Oct 02, 2009 2:07 pm**

by **physics_auth**

physics_auth wrote:betelgeuse1 wrote:You can speak also about bi-elliptical coordinates the basic fact is that you integrate over all the variables you are not interested in. In spherical, if you are interested about the value of your quantity considering only its "radial" coordinate you "average" out the other variables. After integrating over them they become constant parameters and you got it. The fact is easy and quite used in stat-mech. I was somewhat confused about the term used in the problem "radial probability density"... probably because I did not read the term "radial"... AM I AN IDIOT!!!

No you are simply careless ... . Be careful in the official test. To make errors while practising at home doesn't harm as long as you don't repeat them in the real test day!

### Re: small question in QM

Posted: **Sun Oct 04, 2009 4:36 am**

by **betelgeuse1**

I won't repeat them... I'll make brand-new ones....