Solving Ground State of Hydrogen Atom QM Problem

[Beer-monster]

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]

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 $$u$$.

 

[Beer-monster]

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?

 

[Physics Monkey]

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

 

[Beer-monster]

$$-\int^{-1}_{1} {e^(\frac{-ip.z}{\hbar})}.dz$$
Right?
However, I was wondering, if I had made the same substitution but the Sine terms had not canceled (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?

 

[Beer-monster]

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.

 

[Physics Monkey]

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).

 

[Beer-monster]

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?

 

[Beer-monster]

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 canceled 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

 

[Beer-monster]

Anyone see my error, this needs to be in tomorrow

Thanks

 

[NEWO]

hey who started this thread coz i have a piece of work exactly the same question that i am stuck on. I am studying at loughborough uni my name is owen!

 

[Physics Monkey]

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.

 

[NEWO]

beer monkey

if you are doing prof alexanderovs questions for question 1b i get 0.65 (2dp)but I am unsure if that is correct everything seemed to fold out nicely but not sure

 

[Beer-monster]

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

 

[Beer-monster]

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.

 

[Gatts]

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

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