Solving Ground State of Hydrogen Atom QM Problem

[Beer-monster]

Hi guys I'm having a problem with calculations involving the ground state of the hydrogen atom. My main issue comes from the wavefunction of this state: i.e.

\Psi(r) = \frac{1}{\sqrt{\pi a^3}}\exp{\frac{-r}{a}

My main problem seems to come from the fact that this function has no complex element (no i). Which is odd since I thought the wavefunction HAD to be complex also it means there is no complex conjugate. This causes a problem in my calculations for expectation values and probability integrals because there is still a pesky exponential term I can't get rid of which throws a spanner into my working?

One calculation it agravated was finding the probability of the electron being in a classically forbidden zone (i.e with r < a -the Bohr radius)

\int_0^a{\Psi(r)^2}dr

Anyone got any tips, is it my definitions of should I look closer at my algebra?

 

[quasar987]

psi can be all real. Take the infinite well for instace. Or the decaying part in finite wells/barriers.

and what is the problem with evaluating

\int_0^a{\Psi(r)^2}dr = \frac{1}{\pi a^3}\int_0^a e^{-2r/a}dr

?!

 

[Gokul43201]

Beer-monster : You've only written the radial part of the wave function. The complex part comes from the azimuthal term e^{im\phi}.

But in any case, there's no harm in having a real wavenfunction (ie: when m=0). A real number is itself a complex number, and it is its own complex conjugate.

 

[siddharth]

One calculation it agravated was finding the probability of the electron being in a classically forbidden zone (i.e with r < a -the Bohr radius)
\int_0^a{\Psi(r)^2}dr
Anyone got any tips, is it my definitions of should I look closer at my algebra?

Are you not missing a term in your integral? It should be \int {\Psi(r,\theta,\Phi) * \Psi&#039;(r,\theta,\Phi)}dV. So when you simplify that, there will be an additional term

 

[Beer-monster]

Thanks for the quick replies guys, I'll just make this note and then take another crack at the problem.

Goku: I'm working on the ground state on, which I believe is spherically symetric and so has no azimuthal component.

Sidarth: I'm unfamiliar with ther term you have written (assuming \Psi&#039; is the differential of psi?

I thought the calculation of a probability of finding a particle in a region (eg between 0 and a) was

\int_0^a{\Psi(r)^* \Psi(r)}.dr

And if the wavefunction is real doesn't that become

\int_0^a{\Psi(r)^2}dr


Or am I just WAAY off base?

 

[Beer-monster]

tried it again by evaluating.\int_0^a{\Psi(r)^2}dr = \frac{1}{\pi a^3}\int_0^a e^{-2r/a}dr

And I still seem to be getting a very large number which must be wrong, as the probability of the electron existing in the classically forbidden region should be finite but very small, shouldn't it?

However when I integrate the above equation I get

\frac{1}{2\pi a^2}\exp({\frac{-2r}{a})

And when I sub r for the limits of 0 and a I get.

\frac{1}{2\pi a^2}\exp({-2})-1

As a = 5.3x10^-11 I have something small being devided by something very very small, which gives me a VERY big answer. Which makes no sense?

I must have made a stupid error somewhere, but I can't see it? Any guesses

This Quantum stuff is very interesting but becoming increasingly demoralising:(

 

[siddharth]

Sidarth: I'm unfamiliar with ther term you have written (assuming \Psi&#039; is the differential of psi?
I thought the calculation of a probability of finding a particle in a region (eg between 0 and a) was
\int_0^a{\Psi(r)^* \Psi(r)}.dr
And if the wavefunction is real doesn't that become
\int_0^a{\Psi(r)^2}dr
Or am I just WAAY off base?

Why is it \int_0^a{\Psi(r)^* \Psi(r)}.dr. What is the volume element?
For a one dimensional case the probability would be\int_0^a{\Psi(x)^* \Psi(x)}.dx
For a hydrogen atom \Psi is function of r,\theta,\phi. So, the probability will be \int \int \int {\Psi(r,\theta,\phi)^* \Psi(r,\theta,\phi)}.dV
Sorry about my earlier notation. I couldn't find a way to write Psi star in latex.

 

[quasar987]

Siddarth is right. Psi might be spherically symetric but you must nevertheless integrate over a volume. The Born interpretation is that psi*psi.dV gives the probability of finding the electron in a VOLUME dV. And what you're looking for is the probablility that the electron be in a SPHERE of radius a.


At least that's what I think. I might be wrong. I'm only learning this stuff too.

I hope this didn't confuse you more.

 

[Beer-monster]

*slams head into desk*

I am SUCH a moron. I usually write the integral for the 3D case as dxdydz or drd\theta\ d \phi so I thought if I was only considering r I could ignore the other terms. But that would make it 1D not 2D.

I'll try again and integrate between 0 and \frac{4\pi}{3}a^3
Thanks for pointing out I'm an idiot...I have an unfortunate tendency to forget

 

[siddharth]

*slams head into desk*
I am SUCH a moron. I usually write the integral for the 3D case as dxdydz or drd\theta\ d \phi so I thought if I was only considering r I could ignore the other terms. But that would make it 1D not 2D.

Mistakes like this happen to a lot of people .
Also, the Volume element in spherical polar coordinates is NOT dr d\theta d\phi.

 

[Beer-monster]

It's not? *blink*

Oh wait its something like: r^2.dr\sin\theta.d\thetad\phi correct?

And if the wavefunction does not depend on theta or phi

\int_0^\pi {\sin\theta}d\theta \int_0^2\pi {d\phi} = 4\pi

Which should make my integral

4\pi\int_0^a{\Psi(r)^2r^2}dr

Correct?

And on a possibly related subject...am I correct in saying that the probability distribution for a momentum value, is the equation for its expectation value without integration over space i.e

\Psi(r)^*p\Psi(r)

Also should the ground state not be a basis function and therefore the momentum distribution is just hk? Or am I just talking nonsense again?

 

[Gokul43201]

It's not? *blink*
Oh wait its something like: r^2.dr\sin\theta.d\thetad\phi correct?
And if the wavefunction does not depend on theta or phi
\int_0^\pi {\sin\theta}d\theta \int_0^2\pi {d\phi} = 4\pi
Which should make my integral
4\pi\int_0^a{\Psi(r)^2r^2}dr
Correct?

Yes. that's correct, for this case. In general though, the probability density is given by \psi^*\psi~or~|\psi |^2

And on a possibly related subject...am I correct in saying that the probability distribution for a momentum value, is the equation for its expectation value without integration over space i.e
\Psi(r)^*p\Psi(r)

...yes (that will give you the expectation value of the momentum), where "p" is the momentum operator in position space representation...and is given by... ?

Also should the ground state not be a basis function and therefore the momentum distribution is just hk? Or am I just talking nonsense again?

Can't say I have a clue what you're saying here.

 

[Beer-monster]

Goku: Thanks for the quick reply

The momentum operator is -i\hbar\frac{d}{dx} in 1D. However I thought that to get the actual expectation value you had to integrate the \Psi(r)^*p\Psi(r) over all space.

The question is asking for the probability distribution itself, so is that not the above without integration or am I making stuff up?

I read somewhere that basis wavefunction (the ones that all other wavefunctions are a sum of) have no probability distribution for momentum rather they have a defined momentum of hk. However I thought the ground state wavefunction was a basis function, but thinking on it that makes no sense so ignore me :)

Has anyone perchance calculated the answer to the first part. I think it'd be really helpful to have something to compare to.

At the moment I got p = \frac{a^2}{2}\exp{-2} which I got by integrating first by substition of the 2r/a for t in the exponent and then using by parts to work out the integral, using the limits between t=0 and t=2 (since if r=a t=2)

I'm fairly sure that is wrong, but I think I'm getting close.

 

[quasar987]

The question is asking for the probability distribution itself, so is that not the above without integration or am I making stuff up?

I think you're mixing expectation value and probability densities. In the mathematicall theory of probability, if you have a probability density function P(x) [i.e. what the wording of your problem calls a probability distribution], then the probability of getting a result btw a and b is

\int_{a}^b P(x)dx

I'm sure this integral looks familiar to you. In the case of quantum mechanics, then, what is the probability distribution of position? And how is the probability distribution of momentum related to it?

 

[Beer-monster]

I had thought it the probability density (is that the same as distribution?) of position was \psi^*\psi. Thus, following from the classical probability theory above, integrated over all space gives you the probability of finding the particle in space, which is 1, as with normalisation.

I had thought the probability distribution of momentum would therefore be \psi^*p\psi, but that integrated gives the expectation value not the probability, thus I'm lost.

I know the amplitude of the wavefunction is related to probability, however that would give the same value for any operator wouldn't it?

 

[quasar987]

First, yes, probability distribution = probability density. I tried to make a point of clearing that up in my previous thread but apparently it had the opposite effect.

Anyway, which book at you teaching yourself from? (I'm guessing that's what you do)

I'm not one to tell you why that is, but our book (Gasiorozicz [NOT recommended]), says that if you take the Fourier transform of psi, then you get the \phi(p) and that is the "wave function in momentum space" in that |\phi(p)|^2 is the probability distribution of momentum. I don't know if \phi(p) is on a perfectly equal footing as \psi(x) in that for exemple, you could compute expectation values using \phi(p) instead of \psi(x).

For exemple, is the following true?

&lt;K&gt; = \int_{-\infty}^{\infty}\phi^*\left(\frac{p^2}{2m}\right)\phi dp = \int_{-\infty}^{\infty} \psi^*\left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi dx

Probably!

 

[Beer-monster]

Unfortunately no , I'm not teaching myself. I'm an undergraduate student taking a module on quantum physics. We were given a problem sheet for next monday, and though I thought I understood what was being asked, I can't actually get any good answers. So I'm kinda second guessing everything I thought I knew.

The 2 books I'm using to help with the course (along with my profs notes) are A.Levi: Applied Quantum Mechanics. A.I.M Rae, Quantum Mechanics. Both are good in different ways, but I find I can't learn from textbooks only revise.

This is becoming quite soul-crushing

 

[jtbell]

I don't know if \phi(p) is on a perfectly equal footing as \psi(x) in that for exemple, you could compute expectation values using \phi(p) instead of \psi(x).
For exemple, is the following true?
&lt;K&gt; = \int_{-\infty}^{\infty}\phi^*\left(\frac{p^2}{2m}\right)\phi dp = \int_{-\infty}^{\infty} \psi^*\left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi dx

Yes, that is true. You can calculate any expectation value using either \psi(x) or \phi(p), so long as you use the appropriate representation of the operator for that quantity, as you did in your example. The only difference is that it may be easier to grind out the integral in one case or the other.

 

[Beer-monster]

Thanks Quasar, Jtbell. I looked up the releation between momentum and position via a Fourier transform in one of my past textbooks which I used for my first QM modules (Introduction to Quantum Mechanics by AC Phillips, I highly recommend it for the first steps into the subject).
The book also showed the inverse Fourier transform (and I'm glad it did since I recall sucking at FT) to convert \psi(x) to \phi(p) as

\frac{1}{\sqrt{2\pi\hbar}}\int_\infty^\infty {\psi(x)\exp{(-ipx/\hbar)}}dx

I just wanted to check with you guys that I would be correct in converting this to spherical polar co-ords for the ground state Hydrogen atom wavefunction (spherically symetric) as:

\frac{1}{\sqrt{2\pi\hbar}}\int_0^\infty{\psi(r)\exp{(-ipr/\hbar)}r^2}dr

This look okay?

And has anyone tried the calculation for the probability of finding the electron in the classically forbidden zone. I know my answer is wrong, but I'm not sure where I'm slipping up and to know what to aim for would be a big help.

Thanks again guys

 

[quasar987]

For in integral, Mapple gives 1-5e^{-2}.

 

[Beer-monster]

That's uh interesting. Nothing like my answer

 

[Beer-monster]

Okay with regards to the momentum probability density question, I managed to find the true expression for the 3D Fourier Transform as:

\phi(p) = \frac{1}{(2\pi\hbar)^\frac{3}{2}}\int{\exp(\frac{-ipr}{\hbar})\psi(r)}.dV

Unfortunately in spherical polars, I think, this is.

\phi(p)=\frac{A}{(2\pi\hbar)^\frac{3}{2}}\int^0_\infty \int^0_\pi{\exp(\frac{-ipr\cos\theta}{\hbar}-\frac{r}{a})}r^2\sin\theta.dr.d\theta

And I have no idea how to integrate that

Any ideas guys?

 

[quasar987]

why does the r in your exponential becomes rcosO after the transformation in polar?

 

[Beer-monster]

The original equation had the vectors r and p at an angle to each other (with p on the z axis), so their product would be a scalar product of the two which I think can be expressed as pr\cos\theta
I also went with it (because it was in the book) and because I had previously tried expanding a 1D case, by using the changing x to its polar equivalent.
Is it wrong, because that's what really makes it so hard to integrate

 

[quasar987]

I don't know if it's right or wrong. Like I said, I'm just learning this too. However, I just don't see the transition btw your 2 integrals. Also, what is A?!

 

[Beer-monster]

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}}

 

[Beer-monster]

Anyone? I REALLY need this. Please?

 

[Physics Monkey]

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.

 

[Beer-monster]

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]

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.

 

[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 tommorrow

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:

<br /> \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}<br />
<br /> \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}<br />