QM - Hydrogen atom

  • #26
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Oh sorry A is the amplitude of the ground state wavefunction (I put A in my working to simplify the maths - didn't work)

[tex]A = \frac{1}{\sqrt{\pi a^3}}[/tex]
 
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  • #27
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Anyone? I REALLY need this. Please?
 
  • #28
Physics Monkey
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You can do the integral over [tex] \theta [/tex] with a simple substitution. Hint: try to make the thing in the exponential look simple i.e. get rid of the cosine.

Also, your integration bounds are messed up, you have them flipped for some reason. I didn't check all your constants so I can't say if you have those right.
 
  • #29
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I had considered that but I'm not sure which substitution to make. If I subbed the cos for another term say z. I'd get

[tex] -\int^0_\pi {exp(\frac{prz}{\hbar})sin\theta.dz}[/tex]

I think (I've not tried it on paper just my head so far).

However. Even though the new variable of integration is z, can I still treat the [tex]\theta[/tex] as a constant as this was this orginal variable of interest?
 
  • #30
Physics Monkey
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That isn't right. If I call [tex] u = \cos{\theta} [/tex], what is [tex] du[/tex]? I'll give you a hint, it isn't just [tex] du = d\theta [/tex] which is what you appear to have so far.
 
  • #31
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It should be [tex] -sin \theta.du[/tex] since the derivative of cos is -sine. Right? Thats what I was trying to right, but I didn't account for the other sine. So should it be sin^2 instead of sine in the above integral? And can I treat this as a constant when integrating wrt u. If not I'm unsure how to proceed since I'll have two variables then?
 
  • #32
Physics Monkey
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If [tex] du = - \sin{\theta} d\theta [/tex] then you have precisely [tex] du [/tex] in the bottom of your integral already, right? I'm not sure where you are getting this [tex] \sin^2{\theta} [/tex] idea from. The whole point of the substitution is that you transform one seemingly complicated integral over [tex] \theta [/tex] into something like a simple exponential integral in terms of [tex] u [/tex].
 
  • #33
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Sorry my brain burped. :blushing: Integratiuon by substitution is probably my weakest point in calculus.

Anyway, let me show you how it is as I get it, and maybe you can let me know where I'm going wrong.

The tricky theta integral is

[tex]\int^\pi_0 {e^(\frac{-ipCos\theta}{\hbar})Sin\theta}.d\theta[/tex]

If we make the substitution [tex] z = Cos\theta [/tex]

Then [tex] \frac{dz}{d\theta} = -Sin\theta [/tex]

Now as I understand the integration by substitution process, we need to sub [tex]d\theta[/tex] in terms of dz i.e

[tex] d\theta = \frac{-dz}{Sin\theta} [/tex]

This makes our integral.

[tex]\int^\pi_0 {e^(\frac{-ip.z}{\hbar})Sin\theta}.\frac{dz}{Sin\theta}[/tex]

Now would the Sine terms cancel leaving?

[tex] \int^\pi_0 {e^(\frac{-ip.z}{\hbar})dz [/tex]

Which I can integrate, but not 100% sure about the method? Any bugs in the system?
 
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  • #34
Physics Monkey
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You are almost there, you need to write the limits in terms of the new variable. Also, you lost a minus sign somewhere.
 
  • #35
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[tex]-\int^{-1}_{1} {e^(\frac{-ip.z}{\hbar})}.dz[/tex]
Right?
However, I was wondering, if I had made the same substitution but the Sine terms had not cancelled (so ther was still a sine theta in there). Could I have proceeded because I'd now have a a substituted variable and the old variable that need integrating? Could I have treated the theta terms as constant despite theta being the variable of interest?
 
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  • #36
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Right so I solved the theta integral, leaving my r integral as (subbing k for p/h and alpha for 1/a):

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

I tried to solve the integral by parts by subbing u =r and using the standard integrals.

[tex]\int{e^{ax} Sin(bx).dx = \frac{e^{ax}}{a^2+b^2}(aSin(bx)-bCos(bx)[/tex]

and

[tex]\int{e^{ax} Cos(bx).dx = \frac{e^{ax}}{a^2+b^2}(bSin(bx)+aCos(bx)[/tex]

Which I'm not sure about. I get (after fiddling with the constants)

[tex] \frac{2\sqrt{2}}{\pi k(\hbar^3a^3)^{1/2}}[\frac{k\alpha}{(k^2-\alpha^2)^2}] [/tex]

Does this sound about right. Can anyone suggest a way to simplify this more?

Whats the forum stand on checking working? I tried the earlier expectation for the kinetic energy of the hydrogen ground state problem and am very close, except I'm a factor of 2 out and can't see where I lost it.
 
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  • #37
Physics Monkey
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The fourier transform of the ground state goes like [tex] \frac{1}{(k^2 + a^{-2}_0)^2} [/tex], so unfortunately you have made an error in your integration. Quite apart from anything else, your formula can't be right because it is singular at [tex] k = \alpha [/tex] where as the real space wavefunction is non-singular and very well behaved (in other words, you shouldn't get in a singularity in the fourier transform).
 
  • #38
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Right, well the k in the constant and the numerator cancel. So my problem must be the extra alpha. I'll take a look at the working again. However can you tell me if there is anything wrong with using that standard integral to do the transform in r?
 
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  • #39
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Nope still have that pesky alpha, and I feel like I'm so close too.:cry:

Can anyone see where I'm going wrong?

Starting with

[tex] \phi(p) = \frac{2A}{(2\pi/\hbar)^{3/2}}\frac{2\pi}{k}\int^\infty_0{rSin(kr)e^{-\alpha r}}.dr[/tex]

Concentrating on the integral, I used by parts making the allocating the following.

[tex] \frac{dv}{dr} = e^{-\alpha r}Sin(kr)[/tex] thus using the standard integral I mentioned I get [tex] v = \frac{e^{-\alpha r}}{k^2-\alpha^2}(- \alpha Sin(kr)-kCos(kr))[/tex]

[tex]u = r [/tex] thus [tex] \frac{du}{dr} = 1 [/tex]

Subbing this into the by parts formula I get.

[tex] [\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - \frac{- \alpha}{k^2- \alpha^2}\int{e^{-\alpha r}Sin(kr)}.dr -\frac{k}{k^2-\alpha^2}\int{e^{-\alpha r}Cos(kr)}.dr[/tex]

Making the integration gives me three terms, again, using the standard integrals.

[tex][\frac{re^{-\alpha r}}{k^2-\alpha^2}(-\alpha Sin(kr) - kCos(kr))] - [\frac{- \alpha}{k^2-\alpha^2}}(-\alpha Sin(kr) - kCos(kr))] - [\frac{k}{k^2 - \alpha^2}(-\alpha Cos(kr) + kSin(kr))][/tex]

The terms have limits between zero and infinity. The infinity terms cancel as exp(-infinity) is zero. So only the zero substitutions count. As r = 0 the first term is cancelled to zero. And as Sin zero is zero only the cos terms have anny significance. The Cos reduced to 1 and so I'm left with.

[tex]\frac{- \alpha}{k^2- \alpha^2}(\frac{k}{k^2-\alpha^2}) - \frac{k}{k^2-\alpha^2}({\frac{\alpha}{k^2 -\alpha^2})[/tex]

Leaving me with [tex]\frac{2k \alpha}{(k^2-\alpha^2)^2} [/tex]

And thus the same answer as before. The 2 belongs and the k cancels, its just the alpha that shouldn't be there that I can see. I can't see my error (I'm terrible at finding mistakes) can any of you guys (Assuming what I've written makes sense).

Thanks:biggrin:
 
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  • #40
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Anyone see my error, this needs to be in tommorrow

Thanks
 
  • #41
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hey who started this thread coz i have a piece of work exactly the same question that i am stuck on. Im studying at loughborough uni my name is owen!!!
 
  • #42
Physics Monkey
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Beer-monster, you've got the wrong value for the integral, it should be [tex] k^2 + \alpha^2 [/tex] in the denominator, how did you get [tex] -\alpha^2 [/tex]?

You might try to evaluate the integral by breaking the [tex] \sin{kr}[/tex] up into its exponential parts. You should have two integrals to do, one involves [tex] e^{ikr-r/a_0} [/tex] and the other involves [tex] e^{-ik-r/a_0} [/tex] with an important relative minus sign. Each of these integrals is easy to do as it is a standard exponential integration. Once you've done each integral (be very careful with your minus signs), find a common denominator. The form I gave should pop right out.
 
  • #43
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beer monkey

if you are doing prof alexanderovs questions for question 1b i get 0.65 (2dp)but im unsure if that is correct everything seemed to fold out nicely but not sure
 
  • #44
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I've solved the integrals but I get an addition of two [tex] (ik- \alpha)^2 [/tex]. I'm nnot how I can combine these into the form you described monkey? It's late and my brain has fallen asleep
 
  • #45
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Got there in the end. I'd like to thank everyone on the forum who tried to help me with this problem. You've been a huge help in this and hopefully QM more generally.
 
  • #46
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you made a error integrating, i do it by my self and the result is:

[tex]
\phi(p) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 + a^2 p^2 \right)^2}
[/tex]
[tex]
\phi(k) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 +\hbar ^2 a^2 k^2 \right)^2}
[/tex]
 
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