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Angular momentum and hydrogens atom

  1. Jul 5, 2009 #1

    JK423

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    When we solve the hydrogen's atom problem using the Schrodinger equation, we find the eigenstates of energy in the position representation for example. To do that we use the fact that the observables "H, l^2, l_z" have the same eigenstates. My question is about the l_z operator. The z-Axis is a random one. Lets say that the probability distribution has the form 'A' around that axis. Due to the randomness of the z axis, we can resolve the problem using another axis z'. We will obtain the same form 'A' (Of the probability distribution) around the z' axis. So, for the same problem we have the same probability distributions around 2 different axis. Or to put it differently, we have 2 different probabilities for the same point. I understand that only one can ocour at a time, but i cant understand how can we arbitrarily choose an axis and change the electron's prob distribution around the nucleous as we like! Another way to put the question is: How can we 'force' the H operator to have the same eigenstates with the l_z where z is the random axis we chose? By taking a measurement of l_z on that axis?
     
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  3. Jul 5, 2009 #2

    malawi_glenn

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    we have spherical symmetry of H, we can choose whatever axis we want as the z-axis
     
  4. Jul 5, 2009 #3

    JK423

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    Yes i know that. Thats not what i ask.. Say that you choose a random axis to solve the problem. Selecting an axis to solve the problem is crucial cuz all the eigenstates of energy will take their form 'A' around that axis. My question is: If you concider a real atom, which of all the axis possible (Infinite in number) will it 'choose' and how?
     
  5. Jul 5, 2009 #4

    malawi_glenn

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    But since H is spherical symmetric, we can choose any z-axis we want and the result is the same, that was the original question. Now you are asking quite unphysical things..

    Original question:

    "How can we 'force' the H operator to have the same eigenstates with the l_z where z is the random axis we chose?"

    Answer:
    We can since H is spherical symmetric.

    So why is this not what you asked for? It is written black on white.
     
  6. Jul 5, 2009 #5

    JK423

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    I`ll be a little more descriptive because i didnt think you understood me or i miss something..
    Is it?
    Say you choose a random z-axis and solve the problem, and you find one of the eigenstates: Ψz(r,θ)=Α(r) cosθ. At θ=0 it`s Ψ=Α.
    Νow take a new z'-axis at θ=θ0 of the previous one. You solve the problem again and you find ofcourse: Ψz'(r,θ')=Α(r) cosθ' , where θ' is the angle from z'.
    Whats the Ψ on the z' axis for random r?
    Ψz(r,θ=θ0)=Α(r) cosθ0
    and
    Ψz'(r,θ'=0)=Α(r)

    We have two different results for the same problem. Which means that our choise of axis we`ll use, changes the probability distribution in space. So we change the electron`s probability density just because we chose z or z' or something else. It`s not that, whatever axis we`ll use we`ll get the same result! That`s because the pairs H,lz and H,lz' have different eigenstates.

    And my question is:
    I want to "measure" the probability density of the electron in the atom. Which eigenstates of the following pairs ( {H,lz} , {H,lz} ,{H,lz''} , .... )
    will it "choose" to have when i make the measurement?
    Because if the above are correct, each pair gives us different results!

    I hope that i dont say nonsense....
     
  7. Jul 5, 2009 #6

    malawi_glenn

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    but we can not measure the probability density... so there is no fuzzing about it...

    you are asking about nonsense stuff

    The L_z operator is the derivative with azimutal angle, and H is indep of it, thus H will commute with all L_z's you can think of (i.e. z-axis), the wavefunction itself is never an observable in QM, so you don't have to worry.
     
    Last edited: Jul 5, 2009
  8. Jul 5, 2009 #7

    nrqed

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    Very good question.


    The answer depends on the way the atom has been prepared. If is in a symmetrical state (with respect to direction), then the probability distributions calculated using any z axis will be the exact same function (there will be no x, y or z dependence in the probability).

    On the other hand, if there is an angular dependence, then there is a natural choice of z axis (if there is still some axial symmetry). If we use this axis for z, the eigenfunction will be smple. If we use a different z axis, we will get a linear combination of states.
    But in the end, if we evaluate either probability density at a given physical point in space, we will get the same answer.
     
    Last edited: Jul 5, 2009
  9. Jul 5, 2009 #8

    nrqed

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    Your example is unclear. Do you mean theta_0 to be a constant? In that case, after normalizing the two wavefunctions are identical!

    If you mean the angle to be a variable, then there will necessarily be a theta dependence in both solutions, not in just one.

    EDIT: I realized that you meant the function evaluated at a specific angle. Then the answer is that the two functions are identical, if you normalize them.
     
    Last edited: Jul 5, 2009
  10. Jul 5, 2009 #9
    I don't know why no one has given a clear answer to JK423's very straightforward question.

    Let's consider the three p states of the hydrogen atom. The z spin is conventionally +1, 0, and -1. You can recombine these three states to get corresponding states along any arbitrary axis. The easiest ones to calculate are the sum and difference of the +/- states: they give you the zero spin state along the x or y axes respectively. It gets more complicated along arbitrary axes but the idea is the same.

    When you solve a differential equation by separation of variables, you get a system of states which is meant to be used in linear combination to produce any desired state. There is nothing special about the z axis, and it is a common misconception that the hydrogen atom is actually "in" one of these arbitrary states. It is in a linear combination of as many of them as it likes to be in.
     
  11. Jul 5, 2009 #10

    JK423

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    Dont we measure the probability density every time we do a double-slit experiment?
    Besides that, do we -indeed- have different wavefunctions for different axis? When i say different i mean rotated, like the axis. Each wavefunction still keeps its form around it`s axis, but its rotated respectively to the others.

    Yes i mean θ0=constant.
    But they are already normalized!
    I think that you have something different in your mind: You think of a fixed wavefunction in space and you just rotate the z-axis? Because if u think that, it`s not what i mean. I say that if you solve the problem for a random axis you `ll get a wavefunction Ψ. If you re-solve the problem for another axis rotated to the initial, you`ll get the same wavefunction but rotated to the initial, as the new axis.



    I think that i lost you. Can you explain it a little simpler?
     
  12. Jul 5, 2009 #11
    It really has not much to do with quantum mechanics. It's more a fact of how you solve differential equations by separation of variables. If you change coordinate systems you get a different set of solutions. But you can always recreate the original set by linear combinations from the new set. If you solve the wave equation in two dimenstions using rectangular coordinates, you get checkerboard patterns of plane waves. If you use circular coordinates you get bessel functions. But you can recreate the elements of one system using the elements from the other. Resolve the checkerboard waves into plane waves, then integrate these waves around a circle, and you get back the bessel functions.
     
  13. Jul 5, 2009 #12
    Both of these wavefunctions can't be normalized, since they still have the same functional form. I think that's what nrqed was referring to, so that they are equivalent, when you normalize them.
     
  14. Jul 5, 2009 #13

    Pengwuino

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    Sure, but the wavefunction is meaningless by itself. This unrotated [tex]\psi _1 [/tex] and rotated [tex]\psi _2[/tex] , while different, will result in the same probability densities. This whole dependence for the wavefunction on [tex]\theta [/tex] and [tex]\phi[/tex] that results if you make an arbitrary rotation will vanish when you find the probability density because of the symmetry of the system.
     
  15. Jul 5, 2009 #14

    malawi_glenn

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    This is then according to my suggestion =)
     
  16. Jul 5, 2009 #15

    JK423

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    Yes, ofcourse you can recreate any solution by the linear combination of any set of solutions.
    But i think that you `re talking about expressing the same wavefunction using different coordinate systems. Im talking about solving the original problem using a random z-axis, and then re-solve it using another z'-axis rotated to the initial z. You will get a new rotated wavefunction. And that wavefunction is an eigenstate of H so that you wont need create it with a linear combination.
    Since there are infinite possible axis, the wavefunction of a real atom must have random orientation.
    So if i give you an atom in the eigenstate Ψ(r,θ)=Α(r) cosθ, can you draw the probability density?
    Since it`s not a spherical symmetric wavefunction the z-axis is a "special axis" because the wavefunction is symmetric to it.
    Where will the z-axis look at when you draw the wavefunction?
     
  17. Jul 5, 2009 #16

    Pengwuino

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    Exactly! You can even make an infinite number of rotations every-which way and come out with the same distribution even though you have an infinite number of different wavefunctions. I think the OP might be having a problem noting how crucial the symmetry here is and how this would not work in pretty much anything more complicated.
     
  18. Jul 5, 2009 #17

    malawi_glenn

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    Thinking in and understanding symmetries are crucial in physics.

    I will give my Docent Lecture on symmetries =)
     
  19. Jul 5, 2009 #18

    JK423

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    Yes you are right... Since we can create the rotated wavefunction by acting on the unrotated with a unitary operator, if we square both of the wavefunction the operator will vanish and the result will be the same.
    So thats the answer i was looking for!

    Thank you ALL for your help
     
  20. Jul 5, 2009 #19

    JK423

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    Im glad for inspiring you :P
     
  21. Jul 5, 2009 #20

    malawi_glenn

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    I have always been inspired by symmetries.

    You will have the same situation in Gauge field theories, you will choose a gauge of the transformation and then the total Lagrangian will be invariant since only the field modulus square is an observable, not the field itself. (simply speaking)
     
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