Quantum Mechanics - quick integral question.

[binbagsss]

ψnlm, ψnlm', ψnlm'' are energy eigenfunctions of hydrogen, and a function of r, θ and Ø.

(they all have the same n and l, but different m's)

I am given ψ(r,t=0)=ψnlm+ψnlm'+ψnlm'' .

I am after the probability density as a function of θ, and so need to integrate over r and Ø.

I have been introduced to ψnlm=Rnl(r)Θlm(θ) ϕm(Ø)

I am checking my understanding of integration tecniques

First of all, energy eigenfunctions are orthogonal when integrated over all space in r/θ/Ø.

So I integrate over Ø and attain |RnlΘlm|^2+|RnlΘlm'|^2+|RnlΘl m''|^2

Which is fine, so now I need to integrate over r.

∫ ( |RnlΘlm|^2+|RnlΘlm'|^2+|RnlΘl m''|^2 ) r^2 dr. *

But I know that Rnl(r),Θlm(θ), ϕm (Ø) are separately normalised.

So ∫ (Rnl)^2 r^2 dr =1 ( Here Rnl is real ! )

So from * I atttain :

|Θlm|^2+|Θlm'|^2+|Θlm''|^2

which gives me the wrong answer.

If anyone could help shed some light on this, that would be greatly appreciated !

The correct solution uses the orthogonal tecnique, and then continues explicitly , as a pose to using the normalization technique, so I suspect this may be were I am going wrong...

 

[Simon Bridge]

Aside: you can place LaTeX expressions between double-dollar signs
$\renewcommand{\F}{\mathrm{\Phi}} \renewcommand{\T}{\mathrm{\Theta}} \renewcommand{\R}{\mathrm{R}} \renewcommand{\dr}{\mathrm{\;\mathrm{d}r}} \renewcommand{\dth}{\mathrm{\;\mathrm{d}\theta}} \renewcommand{\dph}{\mathrm{\;\mathrm{d}\phi}}$
$$\psi_{nlm}=\R_{nl}(r)\T_{lm}(\theta)\F_m(\phi)$$

You appear to have done:
$$p(\theta)=\int_0^\infty \int_0^\pi |\Psi(\vec r, t=0)|^2 r^2\dph\dr$$ ... the question that immediately springs up is: where does the $r^2$ come from? The volume element has an $r^2 \sin\theta$ in it... I suspect you just need to be more rigorous with your definitions.

 

[binbagsss]

But I only want to integrate over r and ϕ as I am after a p(θ)?

 

[Simon Bridge]

$\renewcommand{\F}{\mathrm{\Phi}} \renewcommand{\T}{\mathrm{\Theta}} \renewcommand{\R}{\mathrm{R}} \renewcommand{\d}{\mathrm{\;\mathrm{d}}} \renewcommand{\dr}{\mathrm{\;\mathrm{d}r}} \renewcommand{\dth}{\mathrm{\;\mathrm{d}\theta}} \renewcommand{\dph}{\mathrm{\;\mathrm{d}\phi}}$

But I only want to integrate over r and ϕ as I am after a p(θ)?

That's right.

Normally $$P_{vol} = \iiint_{vol}|\Psi(\vec r)|\d\tau$$ ... which is to say: the probability of finding the particle in a specified volume is the integral over that volume of the square-modulus of the position-wavefunction.

The volume is indicated by $\tau$ because $V$ is already used in context to represent a potential.
$\d\tau = r^2\sin\theta\dr\dth\dph$ would be the volume element in spherical coordinates.
(Where does that $r^2$ come from? Can you derive this volume element?)

You want to find $p(\theta)$, which is not the same thing.
Can you express $p(\theta)$ in words: what does this bunch of symbols mean?

You seem to be thinking that you should do the above volume integral except for the $\theta$ part ... but (a) that is not what you did above and (b) does that make sense? But I cannot actually tell what you are thinking ... if you don't answer questions, I cannot help you.

 

[binbagsss]

a) That is what I was thinking, and I am struggling to see how this is not what I did.
b) I think I see now, this doesn't make any sense at all as r^2 is not a volume element.

I thought this part wouldn't be relevant to my issue, but due to the volume element I now see it was - the question asked to find the probablity density functin as a function of μ, where μ=cosθ, st θ terms no longer appear:

|dμ|=sinθ

D(μ)dμ = \int\int(ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dμ dø
=> D(μ) = ^{2∏}_{0}\int^{∞}_{0}\int(ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dø.

which was my starting point.

 

[vela]

a) That is what I was thinking, and I am struggling to see how this is not what I did.
b) I think I see now, this doesn't make any sense at all as r^2 is not a volume element.

$r^2\,dr$ has units of volume.

I thought this part wouldn't be relevant to my issue, but due to the volume element I now see it was - the question asked to find the probablity density functin as a function of μ, where μ=cosθ, st θ terms no longer appear:

You can also write the volume element as $d\tau = r^2\,dr\,d(\cos\theta)\,d\phi$ where the limits for $\cos\theta$ run from -1 to 1.

|dμ|=sinθ

You mean $d\mu = \sin\theta\,d\theta$. You can't have a lone differential.

D(μ)dμ = \int\int(ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dμ dø
=> D(μ) = ^{2∏}_{0}\int^{∞}_{0}\int(ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dø.

which was my starting point.

You said your answer was incorrect. How do you know? What are you supposed to get?

 

[binbagsss]

r^{2} dr is not the relevant volume element - here derived from the Jacobian for spherical coordinates.

The solution I am trying to follow uses the method in post 5, where sin/cos \theta and d\theta terms no longer appear.

As in post 1 I thought it was a later step where I went wrong - the orthogonal or normalisation assumptions/knowledge application.

The solution attains a number as the answer, whereas I attain a function of \theta. (the wave functions are all specified).

 

[vela]

r^{2} dr is not the relevant volume element - here derived from the Jacobian for spherical coordinates.

No one said it was. You seemed to have implied that $r^2$ wasn't correct because it didn't have units of volume. I was pointing out the volume element includes $dr$ so that the units work out.

The solution attains a number as the answer, whereas I attain a function of \theta. (the wave functions are all specified).

Did you provide us with the complete problem statement? I don't see how you can get just a number if $l$ isn't specified and you're not integrating over $\theta$.

 

[binbagsss]

I saw that, and the reply was merely aimed to be a correction in recognition of your point.

All n,l,m are specified with the wave functions.

Here's the complete problem statement:

ψ_{210}=\sqrt{\frac{1}{2a^{3}\pi}}\frac{r}{4a}exp^{\frac{-r}{2a}}cos\theta

ψ_{211}=\sqrt{\frac{-1}{a^{3}\pi}}\frac{r}{8a}exp^{\frac{-r}{2a}}sin\thetaexp^{i\phi}

ψ_{21-1}=ψ^{*}_{211}

ψ_{(\underline{r},t=0)}=\frac{1}{\sqrt{3}}(ψ_{210}+ψ_{211}+ψ_{21-1})

The question then is: what is the probability density distribution as a function of μ = cos\theta

 

[vela]

OK, that makes more sense. That combination of spherical harmonics is spherically symmetric, so you should get a uniform distribution as a function of $\cos \theta$. It seems to me the problem may be that you didn't break up the original wave functions properly into $R_{nl}(r)$, $\Theta_{lm}(\theta)$, and $\Phi_m(\phi)$.

 

[binbagsss]

The only obvious mistake in breaking up the wave function that I can see is where the normalization constant should go.
But regardless of how I do this, I attain a function of θ, rather than 1/2.

 

[vela]

Well, we can't tell what you're doing wrong if you don't show us your work.

 

[binbagsss]

D(\mu)d\mu=^{∞}_{0}\int^{2\pi}_{0}\int(\psi_{210}+\psi_{211}+\psi_{21-1})*(\psi_{210}+\psi_{211}+\psi_{21-1}) r^{2}drd\mud\phi

=> D(\mu)= (\psi_{210}+\psi_{211}+\psi_{21-1})*(\psi_{210}+\psi_{211}+\psi_{21-1}) r^{2}drd\phi

Integrate over \phi:

^{2\pi}_{0}\int(\psi_{210}+\psi_{211}+\psi_{21-1})*(\psi_{210}+\psi_{211}+\psi_{21-1}) r^{2}d\phi = |R_{21}\Theta_{10}|^{2} + |R_{21}\Theta_{11}|^{2} + |R_{21}\Theta_{1-1}|^{2}


Integrate over r:

^{∞}_{0}\int = |R_{21}\Theta_{10}|^{2} + |R_{21}\Theta_{11}|^{2} + |R_{21}\Theta_{1-1}|^{2} dr = ^{∞}_{0}\int |R_{21}|^{2}|\Theta_{10}|^{2} + |R_{21}|^{2}|\Theta_{11}|^{2} + |R_{21}|^{2}|\Theta_{1-1}|^{2} dr

= ^{∞}_{0}\int |\Theta_{10}|^{2}+|\Theta_{11}|^{2}+|\Theta_{1-1}|^{2}

(by normalisation of Rnl's)

= cos^{2}\theta + sin^{2}\theta + sin^{2}\theta

 

[vela]

You're not getting the normalization of the $\Theta_{lm}$'s correct.

Take $\psi_{210}$. The constant at the front of
$$\psi_{210} = \sqrt{\frac{1}{8\pi a_0^3}}\left(\frac{r}{2a_0}\right)e^{-r/2a_0}\cos\theta.$$ is a combination of the normalization constants for $R_{21}$, $\Theta_{10}$, and $\Phi_{0}$. You need to separate them out correctly.

 

[binbagsss]

Thanks for your help, I can see were this is going.