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QM - Hydrogen atom

  1. Nov 11, 2005 #1
    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?
     
    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #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]

    ?!
     
    Last edited: Nov 11, 2005
  4. Nov 11, 2005 #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.
     
  5. Nov 12, 2005 #4

    siddharth

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    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
     
    Last edited: Nov 12, 2005
  6. Nov 12, 2005 #5
    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?
     
  7. Nov 12, 2005 #6
    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:(
     
    Last edited: Nov 12, 2005
  8. Nov 12, 2005 #7

    siddharth

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    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.
     
    Last edited: Nov 12, 2005
  9. Nov 12, 2005 #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.
     
    Last edited: Nov 12, 2005
  10. Nov 13, 2005 #9
    *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:
     
  11. Nov 13, 2005 #10

    siddharth

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    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].
     
    Last edited: Nov 13, 2005
  12. Nov 13, 2005 #11
    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?
     
    Last edited: Nov 13, 2005
  13. Nov 13, 2005 #12

    Gokul43201

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

    ...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... ?

    Can't say I have a clue what you're saying here.
     
    Last edited: Nov 13, 2005
  14. Nov 13, 2005 #13
    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.
     
    Last edited: Nov 13, 2005
  15. Nov 14, 2005 #14

    quasar987

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    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?
     
  16. Nov 14, 2005 #15
    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?
     
  17. Nov 14, 2005 #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!
     
  18. Nov 15, 2005 #17
    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:
     
  19. Nov 15, 2005 #18

    jtbell

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    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.
     
  20. Nov 15, 2005 #19
    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:
     
    Last edited: Nov 15, 2005
  21. Nov 15, 2005 #20

    quasar987

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    For in integral, Mapple gives [itex]1-5e^{-2}[/itex].
     
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