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Question about Schrodinger equation, potential and energy

  1. Feb 28, 2005 #1
    Ok, I know that the 1D time-independent Schrodinger equation is [tex]-\frac {\hbar^2} {2m} \frac {d^2 \psi(x)} {dx^2} + V(x) \psi(x) = E \psi(x)[/tex]. Why is it that you can mix potentials and energies in the same equation? For example, if you're saying that V(x) has a constant value, say, [tex]V(x) = V_{0}[/tex] and you're talking about a potential step, then you get something like [tex]k = \frac {\sqrt{2m(V_{0} - E)}} {\hbar}[/tex] for the wavenumber if the total energy is less than the step height. How can you subtract two different quantities? It doesn't make sense to me :(. I guess I didn't think about this much before now, when we've been set a problem where we're given values for E (in eV) and V (in V). Does the potential need to be converted to an energy first? It seems other people have been getting confused by this as well, so it isn't just me.

    If anyone can explain, thanks.
     
  2. jcsd
  3. Feb 28, 2005 #2

    ZapperZ

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    Er... that "V" is potential ENERGY, not potential as in electrostatic potential. Y'know, as in gravitational potential energy is mgh, etc... Haven't you seen something like this when you do Lagrangian mechanics?

    Zz.
     
  4. Feb 28, 2005 #3
    I thought at first it was potential energy, but our lecturer refers to it as the step height.. and then he sets a problem with the step height in volts. I don't do Langrangian mechanics, it was an option module (although I didn't get any options this year and wouldn't have done it anyway.. looks too mathematical).
     
  5. Feb 28, 2005 #4

    ZapperZ

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    But think about it. If you put a charge particle in a potential of V, it has the potential to gain an energy by an amount qV! If it is an electron, the energy it can gain is eV. So just simply having an electrostatic potential is equivalent to a potential energy for any charged particle.

    However, your lecturer is obviously sloppy in setting up k with E-V, while still keeping V as the potential. What is meant here is that you have E - qV.

    Zz.
     
  6. Feb 28, 2005 #5

    dextercioby

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    That's bad.Lagrange & Hamilton formalisms for CM are COMPULSORY PREREQUISITES for QM...

    Daniel.
     
  7. Feb 28, 2005 #6

    Tom Mattson

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    Yes, it's very bad indeed. I wonder how they justify allowing students to take QM without classical Hamiltonian dynamics. How do they expect the students to know what the quantum mechanical Hamiltonian is? :confused:
     
  8. Feb 28, 2005 #7

    Galileo

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    It's not really compulsory to have taken Hamiltonian mechanics.
    Without it, you can certainly learn to do quantum mechanics.

    Everyone can see that [itex]\frac{p^2}{2m}+V[/itex] is an expression for the total energy of the particle. The name 'Hamiltonian' may sound new, but wouldn't be a problem.
    It's just that the 'derivation' of the SE and links to stuff that uses the action principle (which is generally considered somewhat 'advanced') remains ununderstood.
    For an introductory course on QM. You can do without Hamiltonian and Langranian mechanics.
     
  9. Feb 28, 2005 #8

    Tom Mattson

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    I agree that you can learn to do it, but I think it would be more of a hollow algorithm than a real understanding.

    But you wouldn't know why H is a function of the momentum and the position, instead of the velocity and the position. In Physics I, you learn that that expression is [itex]\frac{1}{2}mv^2+V[/itex]. Somewhere, the break has to be made from this convention. If you show that expression you quote without explaining Hamiltonian dynamics, then the switch from (x,v) to (x,p) appears to be without motivation, does it not?

    edit: Does anyone know why LaTeX puts all that whitespace in my expression?
     
    Last edited: Feb 28, 2005
  10. Feb 28, 2005 #9
    In a QM class at this level (I had one) it goes like this:

    There exist a wave function psi such that the partial of psi with respect to position is momentum, and the partial of psi with respect to time is the Energy. Now, energy and momentum are related thusly, so we have the shrodinger equation.

    Q: "Why is partial of psi wrt to x the momentum?"

    A: "Because that is what gives us the right answer".

    To be honest, my class wasn't even this clear. Instead we started with:

    psi = exp( i(kx-wt)) because the double slit particle suggested waves, and then noticed that the derivatives of psi were related according to the shrodinger equation... :cry: :zzz:
     
  11. Mar 1, 2005 #10
    Thanks Zz :smile:. The recommended book for our course (Eisberg and Resnick) does the same) with V, unless they mean potential energy and not just potential.

    Do you mean the Hpsi(x) = Epsi(x) thing? We did get taught that in our QM course.
     
  12. Mar 1, 2005 #11
    No. The whole QM formalism comes from the Hamiltonian by quantizing, and understanding the nature of state space, and dynamics. The Hamiltonian comes from the Lagrangian. Also the Lagrangian formalism carries across to QFT easily too.

    I'm currently in the first year of a physics undergrad course, and feel that they ought to teach us the "Advanced CM" before QM, but they won't - full QM will be taught next year, much to my dismay. Lucky for me, I already know Lagrangian mechanics, and Hamiltonian will be something for me to read this summer.
     
  13. Mar 1, 2005 #12
    This is kind of sad - but I took 3 semesters of QM as an undergrad (1 semester undergrad, and 2 semesters grad) without taking a CM course.
     
  14. Mar 1, 2005 #13

    dextercioby

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    No say,then what the heck did they teach you there...?I think the formulations of Lagrange & Hamilton are called (at least in my country) "analytical mechanics" and they correspond to the first theoretical physics course which is taught in the 3-rd semester (of the academic programme).Newton's formulation is called "classical mechanics" and is taught in the second semester,while QM is taught in the fifth & the 6-th...

    I don't know what system of teaching you got there people,but it's definitely not the one i like.

    Daniel.
     
  15. Mar 1, 2005 #14
    Let me recommend Goldstein's book Classical Mechanics to you. There may better ones now, but I strongly doubt it. At JHU, where I got my Ph.D. in physics 40 years ago, it was the standard for the CM course.
    The book is so good that there were no lectures. The graduate students met once per week and some were called upon to put their solutions to last week's assigned homework problems on the board. A new set of problems for home work was assigned and we went away when we wanted to (15 minutes or two hours later, as we thought best.)
    It was one of the best courses I ever had. Work your way thru that book (It will take at least a hundred hours, unless you are very exceptional, if you do 1/3 of the problems.
    Goldstein's introductory note tells it all. I can't quote it after fourty years, but the essense is "Be ye doers and not just readers." Do 1/3 of his chapter end problems, and you will really know CM. (Don't do just the first three, they get progressively tougher.) You don't need a formal course, but of course a good one, or someone to ask when you get stuck, sure helps. Organize your own informal course with a few friends, if need be.
     
    Last edited: Mar 1, 2005
  16. Mar 1, 2005 #15

    dextercioby

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    In 40 years,there's no other book better than Goldstein in CM...(Landau & Lifschitz is a little shallow).

    Daniel.
     
  17. Mar 1, 2005 #16
    I don't know where juvenal's from, but I'm from the UK at Oxford University, and we don't get advanced CM, unless we choose to do it in the 3rd year.
     
  18. Mar 1, 2005 #17

    dextercioby

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    That sounds really devious...

    Daniel.
     
  19. Mar 1, 2005 #18
    Thanks - but I took a Goldstein course my first term of grad school. That was torturous.
     
  20. Mar 1, 2005 #19
    What do you mean?
     
  21. Mar 1, 2005 #20

    dextercioby

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    What do you get to choose it...?Isn't it COMPULSORY...?And when is the QM course...?

    Daniel.
     
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