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Thread: Schrodinger wave equation

  1. #1 Schrodinger wave equation 
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    I m studying the ch'structure of atom' and I have just reached the topic"quantum mechanical model of atom in which there is a schrodinger wave equation given.In the equation the term x,y and z are given which are the co-ordinate of electron.If the value of the term x,y and z vary the value of the equation should be vary.This is my view.Is it right or wrong?Please explain...thanx!


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  3. #2  
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    Quote Originally Posted by sagarkaran View Post
    I m studying the ch'structure of atom' and I have just reached the topic"quantum mechanical model of atom in which there is a schrodinger wave equation given.In the equation the term x,y and z are given which are the co-ordinate of electron.If the value of the term x,y and z vary the value of the equation should be vary.This is my view.Is it right or wrong?Please explain...thanx!
    I would answer by asking you to think about what the wave equation (any wave equation or, for that matter, any differential equation, partial or otherwise) means. What does psi (the wavefunction) mean (see footnote)? And must it be constant over all space and time, or can it have a spatio-temporal variation?

    Footnote: It may be unfair to ask about the meaning of psi itself, so feel free to consider the meaning of its squared-modulus.


    Last edited by tk421; January 15th, 2013 at 04:15 PM.
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  4. #3  
    mvb
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    Quote Originally Posted by sagarkaran View Post
    I m studying the ch'structure of atom' and I have just reached the topic"quantum mechanical model of atom in which there is a schrodinger wave equation given.In the equation the term x,y and z are given which are the co-ordinate of electron.If the value of the term x,y and z vary the value of the equation should be vary.This is my view.Is it right or wrong?Please explain...thanx!
    I am trying to figure out exactly what is bothering you. Do you have any experience with calculus?
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    No,90 percent I have no idea of calculus.
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    Quote Originally Posted by sagarkaran View Post
    No,90 percent I have no idea of calculus.
    That's unfortunate, because Schrodinger's equation is a partial differential equation. Without a calculus background, it's just gibberish. You'll have to get the requisite background first before you can hope to work with Schrodinger's equation with even the simplest boundary conditions.
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    I m in class 11 and don't know very well that what does wave equation mean,however as all I know about is that wave equation is formulated in order to find the value of amlitude and energy level at any point around nucleus.
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    Does wave equation depends on the coordinate x,y,z axis?
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    Quote Originally Posted by sagarkaran View Post
    Does wave equation depends on the coordinate x,y,z axis?
    What studying have you done? What textbooks have you read? And is this question the result of a homework assignment?
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  10. #9  
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    Quote Originally Posted by sagarkaran View Post
    I m studying the ch'structure of atom' and I have just reached the topic"quantum mechanical model of atom in which there is a schrodinger wave equation given.In the equation the term x,y and z are given which are the co-ordinate of electron.If the value of the term x,y and z vary the value of the equation should be vary.This is my view.Is it right or wrong?Please explain...thanx!
    I presume you are doing this in your own time ? The Schroedinger Equation is not usually part of a year 11 curriculum.
    As tk421 has rightly pointed out already, it will not be possible for you to either understand or solve this equation if you do not have any knowledge of calculus. In fact, even with a good grounding in calculus the SE can be quite tedious to solve manually, especially in three dimensions. My recommendation would be to first study calculus and differential equations in detail, and then come back to the SE.
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  11. #10  
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    The first time I worked the S equation was a 300 level organic chemistry class in college. You definitely need some background in order to hope to understand it.
    "Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us." -Calvin
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  12. #11  
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    While I agree with those that say you are going to have to learn some calculus before you can completely understand the Schrodinger Equation, it is easier to do that work knowing a bit about where you are headed in chemistry. So there is something to be said for knowing roughly what the calculus is doing before you actually learn how to calculate with it. Then you can tackle the mathematics knowing why you want to master it. So, understanding that you will eventually learn about the Schrodinger Equation more deeply, let me see if I can give you an idea about what is going on.

    First, since we aren't trying to figure out how to get all the details, let us work with one dimension only. I will work only in x, and I will use without explaining the solution for the time-dependence. The Schrodinger equation is then

    [ (d2/dx2)psi(x) + V(x) psi(x) -E psi (x) ] = 0

    Now I think that where you were headed is that if d2/dx2 were an algebraic function, you would factor the psi(x)

    [ (d2/dx2) + V(x) - E ] psi(x) = 0

    so that either psi is identically 0 or the factor in brackets is 0. In the second case you learn nothing about psi because the product is zero for any psi whatsoever.

    However, (d2/dx2)psi(x) is not just a constant or even a known function times psi(x). Instead, (d/dx) psi(x) is the slope of a graph of psi(x).

    That means that (d/dx) psi(x) gives you the rate that psi(x) is changing. In a simpler case, if x(t) is your x-position at a given time,

    (d/dt)x(t) is your velocity. So (d/dt)x(t) = v(t). Do it again, and you have (d/dt)(dx(t)/dt = dv/dt which is your acceleration, a(t). Overall,

    using the shorthand

    (d/dt)(dx(t)/dt = (d2/dt2)x(t) = a(t)

    and a(t) can be anything whatsoever. In the case of the Schrodinger Equation, we can rewrite the equation as

    [ (d2/dx2)psi(x) = -V(x) psi(x) + E psi (x) ]

    The term on the left-hand side of this equation changes psi in more complicated ways than multiplying it by a function, so we can't
    conclude either that psi(x) = 0 or that the multiplier of psi vanishes so that we know nothing about psi itself.

    In summary, the Schrodinger Equation can be solved for the form psi(x), and in general that form is a lot more complicated than we would guess if knew about algebraic equations only. For the purpose of seeing what is coming when you study chemistry more deeply, just tell yourself that psi is a function whose form can be determined, and keep on going.

    I will watch for any further questions about this.
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  13. #12  
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    OK, let's use the same longhand notation that mvb uses. Let's also use some simplifying (and not quite correct) assumptions.

    We assume an electron orbiting an atomic nucleus, and we assume that to this electron we can associate a wave function. Now according to N. Bohr, the only allowed angular momenta are integer multiples of . Roughly speaking this is to prevent successive orbits from cancelling the wave function by destructive interference.

    Now generalize: for any subatomic particle with any sort of momentum in 1-dimension write as an operator , say.

    Now write .

    This is the first term in your Hamiltonian. The second term is the potential energy term that depends only on position. It is also an operator in this scheme. So .

    As an operator this acts on the wave function , the result being the allowed energy eigenvalues for our particle - thus

    . That's your Schroedinger equation.

    Note for simplicity I have refrained from working over the complex number field, which would have induced a change of sign, note also that which would have simplified notation somewhat
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    Quote Originally Posted by Guitarist View Post
    Note for simplicity I have refrained from working over the complex number field, which would have induced a change of sign
    Slight nitpick: you need the minus sign over the real numbers also. To see that, just take a free particle [V=0] and use cos(kx) as your wave function.
    Last edited by Guitarist; January 18th, 2013 at 11:34 AM. Reason: Trivial
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  15. #14  
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    OK, fair enough.

    I notice 2 things - mvb took the trouble to develop a reduced form of the time-dependent Schroedinger eqn, whereas my treatment was explicitly time-independent.

    I also notice the OPer fails to acknowledge, let alone show gratitude for, the effort that mvb and I have exerted to help him or her.

    That's forums for you!!
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    Quote Originally Posted by Guitarist View Post
    OK, fair enough.

    I notice 2 things - mvb took the trouble to develop a reduced form of the time-dependent Schroedinger eqn, whereas my treatment was explicitly time-independent.

    I also notice the OPer fails to acknowledge, let alone show gratitude for, the effort that mvb and I have exerted to help him or her.

    That's forums for you!!
    My suspicion is that this was a student who came here looking for a quick answer to a homework assignment, which is why I avoided giving it to him, despite it being easy to do so. He'll probably not show up again soon, if ever. His lack of acknowledgment of your and mvb's generous efforts fits with that hypothesis.
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  17. #16  
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    Ok sir, I obey ur advice.We really need calculus background before SE.Thanks to all !
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