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The Meaning of Schrodingers Equation - In English

  1. Mar 1, 2008 #1
    [SOLVED] The Meaning of Schrodingers Equation - In English

    Ok, so I'm thinking that the Schrodinger's box is not a bad place to be right now.

    This whole Schrodinger equation has got me in turmoil.

    I've searched the world over and can't seem to find an explanation in English about the equation. I actually think it's me being really dense, but I feel better blaming it on the topic :redface:

    Anyways, so I think I get the idea of [tex] PSI^2 [/tex] this merely tells me the probability of finding the particle somewhere in between a and b. I'm thinking of a one dimensional box, with infinite potential energy at a and b and zero PE inside the box.

    So what does [tex] PSI [/tex] tell me? Is [tex] PSI [/tex] relevant? Is it simply the equation for the wave in the box? How does squaring [tex] PSI [/tex] equate to the probability.

    Does anyone have a link for the Schrodinger equation in English for a laymen - non Physics major?

    My apologies for the idiotic question, but I don't seem to be getting it :confused:

    Thanks in advanced for any help!
  2. jcsd
  3. Mar 1, 2008 #2
    In quantum mechanics, the state of any system is represented by a complex function denoted as [tex]\Psi[/tex], also called the wave function. Schrodinger's equation is a differential equation which sets forth rules that wave functions must follow. It does this by expressing the relationship of the value of the wave function with its various partial derivatives. It is a postulate of QM that only wave functions which conform to the relationships set forth by Schrodinger's equation (as well as the specified boundary conditions, i.e. the sides of your box) are considered to be "valid", i.e. they can represent actual physical states.

    Another postulate of QM is that if the wave function for a particle is a function of one or more space coordinates, we can square it to get a probability distribution for those coordinates (actually this is a special case of a similar postulate which lets us get expectation values for any observable quantity). One could come up with some reasons for this, for example since the output of the wave function is complex, it makes sense to multiply the output by its complex conjugate, which will always be a positive real number of the sort we would be expecting if we wanted a probability. Unfortunately arguments like this don't really address the "deeper" question of why the wavefunction relates to the probability this way. However, QM really has nothing to say about such questions; all we know is that when we take a wave function that conforms to these rules, and apply such operations to it, the "right" answers come out (i.e. the experimentally observed energy levels for atoms etc.)
  4. Mar 2, 2008 #3
    Thanks Kyuzo, your explanation was a bit more then I was looking for, but very helpful.

    I think I have found what I was looking for, any comments would be appreciated.

    [tex] \psi (x,t)[/tex] gives the location of a particle as a function of position and time.

    [tex] \psi^2 (x,t)[/tex] gives the probability of the particle being somewhere between a boundary of a and b

    A 1 dimensional square well with infinite potential energy at 0 and L

    [tex] \sqrt{\frac{2}{L}}\int_{0}^{L} \sin^2({\frac{\pi x}{L})} dx[/tex]

    Gives me the probability of of a particle being within the boundary of 0 to L with a wave function of [tex]\sin({\frac{\pi x}{L})}[/tex]

    the value [tex] \sqrt{\frac{2}{L}}[/tex], in this case is the normalization constant.

    The normalization constant is a value which makes [tex] \sqrt{\frac{2}{L}}\int_{0}^{L} \sin^2({\frac{\pi x}{L})} dx = 1[/tex] which tells me that the probability of the particle being between 0 and L is 100%.

    After much searching and reading, this is what I was able to come up with.
    Would anyone be able to confirm my findings?

    Thanks again Kyuzo!
  5. Mar 2, 2008 #4
    Absolutely not. You seem to be getting hung up on the wave function itself having physical significance, which it doesn't; remember that the wave function is complex, so what would it mean for the position of a particle to be represented by a complex number? The wave function carries information about the system only because when we square and integrate it, we can get probabilities for the outcome of making an observation.

    The rest of your post however seems to be on the right track, nice work.
    Last edited: Mar 2, 2008
  6. Mar 3, 2008 #5


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    In classical mechanics, six numbers (position and velocity) represent the state of a point particle at an arbitrary time t, and Newton's second law tells us how to calculate the state at any other time.

    In quantum mechanics, the spatial part of the wave function (i.e. the map [itex]\vec{x}\mapsto\psi(\vec{x},t)[/tex]) represents the state of a point particle at an arbitrary time t, and the Schrödinger equation tells us how to calculate the state at any other time.

    (This is called "the Schrödinger picture". I could also have said that the map [itex]\vec{x}\mapsto\psi(\vec{x},t)[/tex] represents the entire "life" of the particle, from [itex]t=-\infty[/itex] to [itex]t=+\infty[/itex], and the Schrödinger equation tells us how an observer that's translated in time relative to us would describe the same particle. This is called "the Heisenberg picture". Those "pictures" are just two different ways of saying the same thing).

    So to get from the classical description of a point particle to the quantum mechanical description, replace those six numbers with a complex-valued function and Newton's second law with the Schrödinger equation.

    The tricky part is to understand that a point particle in quantum mechanics doesn't have a position like a point particle in classical mechanics. The particle is really "smeared out" over the entire region where the (spatial part of) the wave function is non-zero.
    Last edited: Mar 3, 2008
  7. Mar 3, 2008 #6
    Well, psi represents the wavefunction of a particle, which as you know has no definite position like classical mechanics.
    You can imagine that a quantum particle is spread over space, thus the need for a wavefunction to describe it.....here it would be good if you studied the free particle and wavepackets to see that.
    The fact that we are not sure of the actual position and the use of the wavefunction to describe it is directly connected to the probabilistic nature of QM and the need for averages, i.e. expectation values.

    Now psi can be complex, but psi squared will always be real and as you said is connected with the probability of finding a particle between two points a and b.

    But that is not the only usage of psi- we use psi to calculate expectation values, i.e. mean position, momentum, energy and uncertainties.

    Thus psi is used to give you a lot of information about the system.
    Last edited: Mar 3, 2008
  8. Mar 3, 2008 #7


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    Good post except for your choice of words here. The particle doesn't have an "actual position".
  9. Mar 3, 2008 #8


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    that is correct, since there is no meaning to ask what state a system have before we perform a measuremtn. Therefor a particle only have a actual position after we have measured, causing the wave function to collpase.
  10. Mar 3, 2008 #9


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    This is a matter of interpretation, and as you are well aware, people here often have vigorous discussions about the interpretation of [itex]\psi[/itex] and other things in QM! :biggrin:
  11. Mar 3, 2008 #10
    I am very grateful to all that have replied. I think, I now have a much better understanding of the Schroedinger equation.

    I'll marked this one solved and plug away at some more homework problems.

    Thank you all! :approve:
  12. Mar 3, 2008 #11


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    you can compare shrodinger eq with any probability distribution function also. For example the Gauss distribution etc. The same relations regarding probabilty density, probability, excpectation value etc, is valid even for the S.E.
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