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Application of Schrodinger equation to SHO

  1. Jul 14, 2014 #1
    when Schrodinger equation is applied to SHO only positive value of potential energy changes it to Hermitian polynomial and hence solution is possible but potential energy is positive only when the particle is moving away from the the mean position.The sign of potential is negative when the particle moving towards the mean position hence Schrodinger equation is not applicable because it will not be changed to Hermitian polynomial with negative potential energy. The above observation shows that application of Schrodingef equation is restricted in nature,the fact which is over looked in text books.The potential energy changes sign with change in the direction of motion is conclusively proved below .T and U are the kinetic energy and potential energy .T+U=CONSTANT. dT=-dU, when motion is away from mean position T decreases hence dT is negative which is possible if U positive U=+1/2kx^2. When motion is towards mean position T increases hence dT is positive which is possible only if U is negative U=-1/2KX^2. Any objection be posted with sound logic
     
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  3. Jul 14, 2014 #2

    Simon Bridge

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    ...
    the potential function in the HO schrodinger equation does not depend on the motion of the particle.

    You seem to be confusing the gradient of a function with the function.
    T+U=constant means that dT+dU=0 as you observed.
    So if dT>0, then dU<0 ... so far so good.

    But it is possible for a function to have a negative slope while also having a positive value.
    This is elementary calculus. Therefore it is possible for dT>0 and U>0 even though dU<0.

    Note: classically, the potential energy stored in an extended or compressed spring cannot be negative.


    The main problem with your argument is that it is classical - the concept of the particle moving is not relevant to the application of the schrodinger equation. The classical motion is something that happens on average. QM is a statistical theory so your argument needs to be couched in those terms or, better yet, in terms of an actual calculation using the schodinger equation.

    There is nothing wrong with V(x)<0 in the schrodinger equation.
    In fact it is a very common practice.

    It is possible in the QHO for the particle to be found in the classically forbidden regions.
    This is where the max kinetic energy of the particle is less than the potential at that position.
    The effect is known as tunnelling... the classical argument simply does not apply.

    This effect is routinely covered in QM text books.
     
  4. Jul 15, 2014 #3
    thanks for paying attention to my thread . Do you agree that potential energy of SHO is negative when. the particle is moving towards mean position that is the value is -i/2kx^2 . Is it wrong?The simple way of expressing Schrodinger equation is Hamiltonian operator psi+constant(E-U)PSI=0 If the value of potential energy U negative the equation can not be changed to Hermitian polynomial hence no solution is possible.It is quite simple.As the direction of motion of particle changes it is simple logic that there should be change in sign of potential energy which can be proved analytically also.Schrodinger equation is not such an ideal equation that it is flawless and it is sacrilege to point to its defect.Your defence is not based on plane logic.why U can not be negative when there is two way motion of the particle.Negative value is rejected only because it does not lead to solution only positive value is retained because it suits but motion of particle is not restricted because particle with restricted motion is not SHO.The application of Schrodinger equation to hydrogen atom leads to negative value pf energy but Wheeler was not comfortable with the negative value.The negative value was justified by taking as energy of formation of hydrogen atom from electron and proton. If it were so the hydrogen atom should stable but actually it is not so
     
  5. Jul 15, 2014 #4
    that the particle can go beyond extreme position has nothing to do with the present problem .My contention is that potential energy of particle changes sign with change in direction which leads to conditional application of Sccrodinger equation How do you refute it. it is basic problem which is being intentionally ignored to justify Schrodinger equation without any sound logic
     
  6. Jul 15, 2014 #5

    jtbell

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    No, it means that dU is positive. U does not necessarily have to be positive, although in this case (the SHO) it is indeed positive, U = +(1/2)kx2. When the motion is away from the mean position (x=0), the magnitude of x increases, therefore U also increases, i.e. dU is positive.

    No, it means that dU is negative. Using U as given above, when the motion is towards the mean position (x=0), the magnitude of x decreases, therefore U also decreases, i.e. dU is negative.
     
  7. Jul 16, 2014 #6

    Simon Bridge

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    No I do not.
    I said in my last post, very clearly, that the potential does not depend on the particle motion.
    Please reread my last post.

    This is also not correct.
    Consider the potential: U=-V: -a/2 < x < a/2, U=0 else : V>0.
    This is a finite square well - you can look it up.
    Notice that the potential is negative in some region and yet solutions exist.
    If I put U=∞ else, then it would be an infinite square well with a negative potential - that also has solutions.

    Which "Wheeler" - do you have a reference for this statement?

    Note:
    (1) the negative potential for the hydrogen atom has nothing to do with quantum theory - in fact it predates it - it is entirely classical and arises from the definition of potential energy. The negative energy states for hydrogen, by itself, comes from the negative potential function. This is another example of where negative potentials still get solutions.

    (2) The happiness or otherwise of someone is irrelevant: Nature does not care about what makes people happy.

    That's just not true. Where are you getting this from?
    Please provide a reference so we can see the context.
    Thank you.
     
  8. Jul 16, 2014 #7

    Simon Bridge

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    This contention is not correct.
    Where did you get this idea from?

    You have had the error of this idea explained to you at length in another thread:
    https://www.physicsforums.com/showthread.php?t=730150
    ... in that thread you also consistently confused U and dU: these are different.

    I just did - see post #2.
    There I concentrated on the mistakes you make in your own logic.

    The harmonic oscillator potential does not depend on the direction of motion of the particle - this is as true for QM as it is in classical physics. Since your own argument hinges on PE changing sign, the fact that it does not is sufficient to refute your argument. That is sound logic.

    But I don't expect you to believe me just because i say so - I'll have to demonstrate that the PE is independent of velocity:
    Please see
    http://faculty.wwu.edu/vawter/PhysicsNet/Topics/SHM/MassOnSpring.html [Broken]
    ... for example.

    Notice that the kinetic energy ##T=\frac{1}{2}mv^2## ... does not depend on the direction of the veocity? If U did depend on the direction of the velocity, then T+U would not be a constant, violating the conservation of energy law that your own argument relies on.

    That was all classical - in QM, the "energy levels" you are probably familiar with are the energies ofr "stationary states". Stationary means "not moving".
    It flows that a particle in a stationary state is not moving.
    So even if you were correct, which you are not, then the potential still won't change sign because the particle is not moving.

    Lastly - you have yet to come up with an experiment that will show the effect of this proposed sign-change. It would have to be an experiment that has not been performed so what is it? I'll do it and, if you are correct, I'll get a Nobel Prize at least!

    There is no problem so there is nothing to ignore.
    That is sound logic.

    It is very easy for anyone to conduct an experiment that will verify that the potential energy does not change sign for a horizontal mass on a spring...so why not do the experiment yourself?

    This is the bottom line in physics - everything has to boil down to some experiment that must be published in such a way that anyone can, in principle, duplicate it. This makes it very difficult to cover up flaws in a theory.
     
    Last edited by a moderator: May 6, 2017
  9. Jul 16, 2014 #8
    Try to look for the MWI for negativ energy and worlds spliting and uniting and for back in time energy and there mullty option of position
     
  10. Jul 16, 2014 #9
    thank you all for your learned views on the simple thread.My views are based on well defined rule s . The following discussion will also prove that there is change in sign of potential energy with change in direction of motion
    The concept of potential energy is based on the the fundamental principle not cooked by me
    dU=-W u and w are potential energy and work. When the particle is moving away from mean position work has to done on the particle because it is moving against the force or energy is being spent hence w is negative which leads to positive potential energy.the negative sign to work has been assigned according to convention and i have not broken the established rule . When the particle is moving towards the mean position the particle is moving in the direction of force therefore work is being done by the particle and hence w is positive by convention which leads to negative value of potential energy U. Have i broken any established rule to arrive at the above conclusion. Please point out if there is any.Again there is no valid solution of Schrodinger equation for negative value of potential energy.We can not assign any arbitrary sign to potential energy. there is certain rule.The sign of potential is very important while solving Schrodinger equation.Why bring in irrelevent dichtomy of classical mechanics and quantum mechanics to discuss very basic and fundamental problem
     
  11. Jul 16, 2014 #10

    Doc Al

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    No, it leads to a negative change in potential energy. As the particle returns to equilibrium, the potential energy becomes lower. So what?
     
  12. Jul 16, 2014 #11
    As i far remember zero energy isnt the limit for negative change in potential energy , and there thing like tuneling that some how consernd
     
  13. Jul 16, 2014 #12

    jtbell

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    As Simon pointed out, this is the OP's second thread on this topic. Since he apparently refuses to accept corrections, both threads are now closed.
     
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