# QM - Hydrogen atom

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

$$A = \frac{1}{\sqrt{\pi a^3}}$$

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Anyone? I REALLY need this. Please?

Physics Monkey
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You can do the integral over $$\theta$$ 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.

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

$$-\int^0_\pi {exp(\frac{prz}{\hbar})sin\theta.dz}$$

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 $$\theta$$ as a constant as this was this orginal variable of interest?

Physics Monkey
Homework Helper
That isn't right. If I call $$u = \cos{\theta}$$, what is $$du$$? I'll give you a hint, it isn't just $$du = d\theta$$ which is what you appear to have so far.

It should be $$-sin \theta.du$$ 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?

Physics Monkey
Homework Helper
If $$du = - \sin{\theta} d\theta$$ then you have precisely $$du$$ in the bottom of your integral already, right? I'm not sure where you are getting this $$\sin^2{\theta}$$ idea from. The whole point of the substitution is that you transform one seemingly complicated integral over $$\theta$$ into something like a simple exponential integral in terms of $$u$$.

Sorry my brain burped. 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

$$\int^\pi_0 {e^(\frac{-ipCos\theta}{\hbar})Sin\theta}.d\theta$$

If we make the substitution $$z = Cos\theta$$

Then $$\frac{dz}{d\theta} = -Sin\theta$$

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

$$d\theta = \frac{-dz}{Sin\theta}$$

This makes our integral.

$$\int^\pi_0 {e^(\frac{-ip.z}{\hbar})Sin\theta}.\frac{dz}{Sin\theta}$$

Now would the Sine terms cancel leaving?

$$\int^\pi_0 {e^(\frac{-ip.z}{\hbar})dz$$

Which I can integrate, but not 100% sure about the method? Any bugs in the system?

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Physics Monkey
Homework Helper
You are almost there, you need to write the limits in terms of the new variable. Also, you lost a minus sign somewhere.

$$-\int^{-1}_{1} {e^(\frac{-ip.z}{\hbar})}.dz$$
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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Right so I solved the theta integral, leaving my r integral as (subbing k for p/h and alpha for 1/a):

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

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

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

and

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

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

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

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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Physics Monkey
Homework Helper
The fourier transform of the ground state goes like $$\frac{1}{(k^2 + a^{-2}_0)^2}$$, 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 $$k = \alpha$$ 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).

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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Nope still have that pesky alpha, and I feel like I'm so close too.

Can anyone see where I'm going wrong?

Starting with

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

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

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

$$u = r$$ thus $$\frac{du}{dr} = 1$$

Subbing this into the by parts formula I get.

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

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

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

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.

$$\frac{- \alpha}{k^2- \alpha^2}(\frac{k}{k^2-\alpha^2}) - \frac{k}{k^2-\alpha^2}({\frac{\alpha}{k^2 -\alpha^2})$$

Leaving me with $$\frac{2k \alpha}{(k^2-\alpha^2)^2}$$

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

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Anyone see my error, this needs to be in tommorrow

Thanks

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!!!

Physics Monkey
Homework Helper
Beer-monster, you've got the wrong value for the integral, it should be $$k^2 + \alpha^2$$ in the denominator, how did you get $$-\alpha^2$$?

You might try to evaluate the integral by breaking the $$\sin{kr}$$ up into its exponential parts. You should have two integrals to do, one involves $$e^{ikr-r/a_0}$$ and the other involves $$e^{-ik-r/a_0}$$ 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.

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

I've solved the integrals but I get an addition of two $$(ik- \alpha)^2$$. I'm nnot how I can combine these into the form you described monkey? It's late and my brain has fallen asleep

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.

you made a error integrating, i do it by my self and the result is:

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

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