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Normalization of wavefunctions

  1. Jul 11, 2014 #1

    dyn

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    Hi
    I have been working my way through some past papers and then checking the solutions but I am confused about the following. One question asked for the normalisation constant for the following wavefunction ψ( r, θ , ø ) = Aexp(-r/R) where R is a constant. The solution requires a triple volume integral using dV=r^2sinθdrdθdø. The 2nd question asks to normalise exp(imø) but this time the integral is just done over dø from 0 to 2∏.
    The 1st wavefunction only involves r but I have to do a triple integral. The 2nd only involves ø but I just need a single integral. I'm confused !
     
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  3. Jul 11, 2014 #2

    strangerep

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    You have to integrate over the domain on which the wavefunction is defined. In your 1st case, it's a 3D domain. In the 2nd case it's a 1D domain (actually, just a circle).
     
  4. Jul 11, 2014 #3

    dyn

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    In the first case I can see that it is a 3-D domain as the wavefunction is given as ψ ( r , θ , ø ) even though the function only depends on r. But in the 2nd case the function is just given as exp(imø) not in the form ψ ( ) so how would know if the domain is 3-D or 1-D ?
     
  5. Jul 12, 2014 #4

    strangerep

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    If the question is well-posed, then it should be clear from the context. If it just says ##e^{im\phi}##, with no other information, then it's reasonable to assume it depends only on ##\phi##. I can't say for sure without seeing the wording (and context) of the question(s).

    Also, if you try normalizing ##e^{im\phi}## over an unrestricted 3D domain, you won't get a sensible answer since the integral diverges.
     
  6. Jul 14, 2014 #5

    dyn

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    I've just done another question which asked to determine the constant A in the wavefunction
    ψ ( r ) = Aexp(-r/a). The solution involved the triple integral over r ,θ , ø. But if the function is given as a function of only one variable how do I know to do the triple integral ? Whereas when given exp(imø) which again is only one variable I only do a single integral ?
     
  7. Jul 15, 2014 #6

    strangerep

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    I can't offer any further answer without seeing the full questions. Can you post a link to them?
     
  8. Jul 15, 2014 #7

    dyn

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    Hi,
    I can't post a link to them but I will give the question as best I can.( I don't know how to post complicated equations , sorry )

    The question starts by giving the time-independent Schrodinger equation for a particle in 3-D where ψ = ψ ( vector r ). The question then asks to show that for an electron in a hydrogen atom , the spherically symmetric solutions satisfy an equation which only involves the r coordinate. ( the Laplacian operator in spherical polar coordinates is given as a hint ).
    The next part then asks to show that the wavefunction ψ ( scalar r ) = A exp(-r/a) is a solution to the equation I had previously calculated.
    I don't understand how to know whether to do the triple or single integral as the function only involves r ; but with the exp(imø) it was only a single integral.
    Sorry if i'm not making much sense.
     
  9. Jul 15, 2014 #8

    PeterDonis

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  10. Jul 15, 2014 #9

    jtbell

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    A hydrogen atom is a three-dimensional object, so it is described by a three-dimensional wave function. The fact that the wave function depends only on r (for the specific state that you're dealing with) means that the probability distribution is spherically symmetric; it doesn't turn the atom into a one-dimensional object.

    As a classical analogy, consider a solid sphere of uniform density. All you need to describe it is the density ρ and the radius R. But if you want to find the mass of the sphere, you need to integrate the density over a three-dimensional volume. This would give you the result M = (4/3)πR3ρ, i.e. the volume of a sphere of radius R, times the density.

    As for the ##e^{im\phi}##, does the problem statement not give any context at all for it? What is it supposed to be describing? :confused:

    One possibility that comes to my mind is that the three-dimensional hydrogen wave function is a product of three one-dimensional functions: ##\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)##. When you normalize it, the three-dimensional integral separates into a product of three one-dimensional integrals. It's possible to normalize each of them individually so the overall integral is 1·1·1 = 1.
     
  11. Jul 15, 2014 #10

    dyn

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    Thanks for that. The exp(imø) problem is for the eigenfunction of the z-component of angular momentum with eigenvalue mh/2∏. The normalization integral is a single integral from 0 to 2∏. I understand that this would be impossible to normalise with r ranging from 0 to ∞ but it don't understand why this is a single integral when the other ones are triple integrals.
     
  12. Jul 16, 2014 #11

    dyn

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    The ##e^{im\phi}## I mentioned is the ##\Phi(\phi) you mention above so as you say it would only involve a 1-D integral so would the R(r) integral. But there lies my confusion ! Both ψ (r) and R(r) are functions of a single coordinate but the ψ (r ) is a 3-D integral and R(r) is a 1-D integral
     
  13. Jul 16, 2014 #12

    dyn

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    Sorry the above post didn't come out right. I was trying LateX for the first time. Only got it half right !
     
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