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Quantum mechanics -harmonic oscillators watch

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    Say we have a 1D harmonic oscillator with potential V = \frac{1}{2} m \omega_c^{2} x^2 (classical frequency). Let the wavefunction of a particle be described by \psi_0 = A \exp(-bx^2).

    How do I find the probability of finding the particle outside of the 'classically allowed region'? I figured the classical limit is basically the amplitude a of the particle's SHM, so I integrate the wavefunction squared from a to infinite (and multiply by 2 for symmetry). Not quite sure how to find a, though.

    Thanks for any help !
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    (Original post by trm90)
    Say we have a 1D harmonic oscillator with potential V = \frac{1}{2} m \omega_c^{2} x^2 (classical frequency). Let the wavefunction of a particle be described by \psi_0 = A \exp(-bx^2).

    How do I find the probability of finding the particle outside of the 'classically allowed region'? I figured the classical limit is basically the amplitude a of the particle's SHM, so I integrate the wavefunction squared from a to infinite (and multiply by 2 for symmetry). Not quite sure how to find a, though.

    Thanks for any help !
    Find A using the normalisation condition (probability of particle existing somewhere between -infty and +infty sums to 1), i.e.

     <\psi_0|\psi_0^{*}> = 1

    Otherwise I think the method you suggest is correct!
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    (Original post by Prime Suspect)
    Find A using the normalisation condition (probability of particle existing somewhere between -infty and +infty sums to 1), i.e.

     <\psi_0|\psi_0^{*}> = 1

    Otherwise I think the method you suggest is correct!
    Oh, so the wavefunction amplitude is pretty much equivalent to the classical amplitude? In that case that would make sense. Thanks very muc h!
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    (Original post by trm90)
    Oh, so the wavefunction amplitude is pretty much equivalent to the classical amplitude? In that case that would make sense. Thanks very muc h!
    Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

    Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

     E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

    The classical harmonic oscillator has a constant total energy, which is given by

     E_{tot} = \dfrac{mw_c^2a^2}{2}

    hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

     \dfrac{mw_c^2a^2}{2} =  <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!
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    (Original post by Prime Suspect)
    Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

    Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

     E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

    The classical harmonic oscillator has a constant total energy, which is given by

     E_{tot} = \dfrac{mw_c^2a^2}{2}

    hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

     \dfrac{mw_c^2a^2}{2} =  <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!
    Thanks very much, and the method seems more than viable. I'm going to give this a go on the train home and will let you know if I work it out later :-)
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    (Original post by Prime Suspect)
    Ah sorry I didn't read your first post properly, got the a's confused - thought you wanted to find A not a...

    Unfortunately its not true that A = a; instead I think you want to say that the expected quantum value for energy (the average value, or observed value) should equal the observed energy of the classical harmonic oscillator, i.e.

     E_{av} = <\psi_0|\hat{H}|\psi_0^{*}>

    The classical harmonic oscillator has a constant total energy, which is given by

     E_{tot} = \dfrac{mw_c^2a^2}{2}

    hence I think this total energy can be equated to the expected energy of the quantum system. Therefore I think that the equation

     \dfrac{mw_c^2a^2}{2} =  <\psi_0|\hat{H}|\psi_0^{*}> should hold. From this I think you should be able to determine a and hence solve the problem... I hope!
    EDIT: Argh, messed things up again!

    If my Hamiltonian operator is of the form \dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).
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    (Original post by trm90)
    EDIT: Argh, messed things up again!

    If my Hamiltonian operator is of the form \dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).
    Use integration by parts and use that

     \displaystyle \int xe^{-2bx^2} dx = \dfrac{-1}{4b}e^{-2bx^2}

    (presence of the derivative)

    I think that should work?

    EDIT

    Perhaps my method isn't correct then if you're not supposed to need to evaluate any integrals? Maybe instead you're supposed to consider energies, as for any energy  E \neq \dfrac{mw_c^2a^2}{2} the quantum oscillator is in a region that's classically forbidden... not sure how to use that info though!
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    (Original post by trm90)
    EDIT: Argh, messed things up again!

    If my Hamiltonian operator is of the form \dfrac{-h'^2}{2m} \dfrac{d^2}{dx^2} + V(x) then I'm not sure how I'm supposed to about any of the integrals, as I'll get two integrals with x^2 exp(...x^2) terms and I'm not told how to evaluate those (my lecturer specifically said that all standard integrals will be provided and that we won't have to spend even 10 seconds trying to evaluate them, and only the answer to an integral of the form exp(-2bx^2) is provided).
    Surely you can use the fact that  |\psi_0> is an eigenstate of the Hamiltonian so you don't need to do any integrals.
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    EDIT: Okay, I evaluated the Hamiltonian and got:

    E = \dfrac{h'^2 b}{m} - 4b^2 x^2 + \dfrac{1}{2}m \omega_c^{2} x^2

    I equated this with the energy \dfrac{1}{2} m \omega_c^{2} x^2 and got:

    x = \sqrt{\dfrac{h'^2}{4mb}}

    where b = \dfrac{m \omega}{2h'}

    does that make sense?
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    (Original post by trm90)
    EDIT: Okay, I evaluated the Hamiltonian and got:

    E = \dfrac{h'^2 b}{m} - 4b^2 x^2 + \dfrac{1}{2}m \omega_c^{2} x^2

    I equated this with the energy \dfrac{1}{2} m \omega_c^{2} x^2 and got:

    x = \sqrt{\dfrac{h'^2}{4mb}}

    where b = \dfrac{m \omega}{2h'}

    does that make sense?
    Sunelir is right - you know the ground state energy of the quantum oscillator is given by

     <\psi_0|\hat{H}|\psi_0> =  \dfrac{\bar{h}w_c}{2}

    because  \psi_0 is an eigenstate of the hamiltonian this must hold. So you equate this to the classical energy to find a - sorry for making you evaluate the integrals, I am definitely a bit rusty on the old quantum mechanics!
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    (Original post by Prime Suspect)
    Sunelir is right - you know the ground state energy of the quantum oscillator is given by

     <\psi_0|\hat{H}|\psi_0> =  \dfrac{\bar{h}w_c}{2}

    because  \psi_0 is an eigenstate of the hamiltonian this must hold. So you equate this to the classical energy to find a - sorry for making you evaluate the integrals, I am definitely a bit rusty on the old quantum mechanics!
    I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?

    But I think I can take it from here, thanks guys!
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    (Original post by trm90)
    I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?
    Well the label on psi was 0 and not n.
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    (Original post by trm90)
    I see... this is a silly question, but I was supposed to know that psi was the ground state wavefunction right :o: ?

    But I think I can take it from here, thanks guys!
    Yeah you can safely assume it is given the 0 subscript... I totally missed this at first as well!
 
 
 
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