QM - Hydrogen atom

  • #1
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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.

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

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)

[tex] \int_0^a{\Psi(r)^2}dr [/tex]

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

  • #2
quasar987
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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

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

?!
 
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  • #3
Gokul43201
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Beer-monster : You've only written the radial part of the wave function. The complex part comes from the azimuthal term [itex]e^{im\phi}[/itex].

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.
 
  • #4
siddharth
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Beer-monster said:
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)
[tex] \int_0^a{\Psi(r)^2}dr [/tex]
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 [tex] \int {\Psi(r,\theta,\Phi) * \Psi'(r,\theta,\Phi)}dV [/tex]. So when you simplify that, there will be an additional term
 
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  • #5
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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 [tex]\Psi'[/tex] 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

[tex] \int_0^a{\Psi(r)^* \Psi(r)}.dr [/tex]

And if the wavefunction is real doesn't that become

[tex] \int_0^a{\Psi(r)^2}dr [/tex]


Or am I just WAAY off base?
 
  • #6
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tried it again by evaluating.[tex] \int_0^a{\Psi(r)^2}dr = \frac{1}{\pi a^3}\int_0^a e^{-2r/a}dr[/tex]

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

[tex] \frac{1}{2\pi a^2}\exp({\frac{-2r}{a})[/tex]

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

[tex]\frac{1}{2\pi a^2}\exp({-2})-1[/tex]

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:(
 
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  • #7
siddharth
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Beer-monster said:
Sidarth: I'm unfamiliar with ther term you have written (assuming [tex]\Psi'[/tex] 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
[tex] \int_0^a{\Psi(r)^* \Psi(r)}.dr [/tex]
And if the wavefunction is real doesn't that become
[tex] \int_0^a{\Psi(r)^2}dr [/tex]
Or am I just WAAY off base?

Why is it [tex] \int_0^a{\Psi(r)^* \Psi(r)}.dr [/tex]. What is the volume element?
For a one dimensional case the probability would be[tex] \int_0^a{\Psi(x)^* \Psi(x)}.dx [/tex]
For a hydrogen atom [tex] \Psi [/tex] is function of [tex] r,\theta,\phi [/tex]. So, the probability will be [tex] \int \int \int {\Psi(r,\theta,\phi)^* \Psi(r,\theta,\phi)}.dV [/tex]
Sorry about my earlier notation. I couldn't find a way to write Psi star in latex.
 
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  • #8
quasar987
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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.
 
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  • #9
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*slams head into desk*

I am SUCH a moron. I usually write the integral for the 3D case as dxdydz or [tex]drd\theta\ d \phi[/tex] 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 [tex]\frac{4\pi}{3}a^3[/tex]
Thanks for pointing out I'm an idiot....I have an unfortunate tendency to forget:wink:
 
  • #10
siddharth
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Beer-monster said:
*slams head into desk*
I am SUCH a moron. I usually write the integral for the 3D case as dxdydz or [tex]drd\theta\ d \phi[/tex] 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:smile: .
Also, the Volume element in spherical polar coordinates is NOT [tex] dr d\theta d\phi [/tex].
 
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  • #11
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It's not? *blink*

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

And if the wavefunction does not depend on theta or phi

[tex]\int_0^\pi {\sin\theta}d\theta \int_0^2\pi {d\phi} = 4\pi[/tex]

Which should make my integral

[tex] 4\pi\int_0^a{\Psi(r)^2r^2}dr[/tex]

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

[tex]\Psi(r)^*p\Psi(r) [/tex]

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?
 
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  • #12
Gokul43201
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Beer-monster said:
It's not? *blink*
Oh wait its something like: [tex] r^2.dr\sin\theta.d\thetad\phi[/tex] correct?
And if the wavefunction does not depend on theta or phi
[tex]\int_0^\pi {\sin\theta}d\theta \int_0^2\pi {d\phi} = 4\pi[/tex]
Which should make my integral
[tex] 4\pi\int_0^a{\Psi(r)^2r^2}dr[/tex]
Correct?
Yes. that's correct, for this case. In general though, the probability density is given by [itex]\psi^*\psi~or~|\psi |^2[/itex]

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
[tex]\Psi(r)^*p\Psi(r) [/tex]
...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.
 
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  • #13
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Goku: Thanks for the quick reply

The momentum operator is [tex] -i\hbar\frac{d}{dx} [/tex] in 1D. However I thought that to get the actual expectation value you had to integrate the [tex]\Psi(r)^*p\Psi(r)[/tex] 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 [tex] p = \frac{a^2}{2}\exp{-2} [/tex] 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.
 
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  • #14
quasar987
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Beer-monster said:
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

[tex]\int_{a}^b P(x)dx[/tex]

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?
 
  • #15
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I had thought it the probability density (is that the same as distribution?) of position was [tex] \psi^*\psi[/tex]. 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 [tex]\psi^*p\psi[/tex], 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?
 
  • #16
quasar987
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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. :rofl:

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 [itex]\phi(p)[/itex] and that is the "wave function in momentum space" in that [itex]|\phi(p)|^2[/itex] is the probability distribution of momentum. I don't know if [itex]\phi(p)[/itex] is on a perfectly equal footing as [itex]\psi(x)[/itex] in that for exemple, you could compute expectation values using [itex]\phi(p)[/itex] instead of [itex]\psi(x)[/itex].

For exemple, is the following true?

[tex]<K> = \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[/tex]

Probably!
 
  • #17
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Unfortunately no:blushing: , 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:cry:
 
  • #18
jtbell
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quasar987 said:
I don't know if [itex]\phi(p)[/itex] is on a perfectly equal footing as [itex]\psi(x)[/itex] in that for exemple, you could compute expectation values using [itex]\phi(p)[/itex] instead of [itex]\psi(x)[/itex].
For exemple, is the following true?
[tex]<K> = \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[/tex]
Yes, that is true. You can calculate any expectation value using either [itex]\psi(x)[/itex] or [itex]\phi(p)[/itex], 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.
 
  • #19
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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 [tex]\psi(x)[/tex] to [tex]\phi(p)[/tex] as

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

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:

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

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.:smile:

Thanks again guys:biggrin:
 
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  • #20
quasar987
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For in integral, Mapple gives [itex]1-5e^{-2}[/itex].
 
  • #21
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:bugeye: That's uh interesting. Nothing like my answer
 
  • #22
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Okay with regards to the momentum probability density question, I managed to find the true expression for the 3D Fourier Transform as:

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

Unfortunately in spherical polars, I think, this is.

[tex] \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[/tex]

And I have no idea how to integrate that:cry:

Any ideas guys?
 
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  • #23
quasar987
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why does the r in your exponential becomes rcosO after the transformation in polar?
 
  • #24
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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 [tex] pr\cos\theta [/tex]
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 thats what really makes it so hard to integrate
 
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  • #25
quasar987
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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?!
 

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