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Schrodinger equation problem : part 2

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data

    b)Use the equations [tex]
    a_{-}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k/(2m)^(1/2)}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0
    [/tex] and [tex]
    a_{+}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x+\left(\frac{1}{2k/(2m)^(1/2)}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0
    [/tex] to express as a linear combination of the lowering and raising operators a_ and a+, and then use the operator methods to show that x[itex]
    \varphi_0(x)
    [/itex] is the(not normalized) wave function for the first excited state;

    c) Use the properties of the raising operator to argue that the wave function for the nth energy level, with E_n=(n+1/2)*omega*[tex]\hbar[/tex], for the oscillator is a polynomial multiplied by a gaussian exp(-c*x^2), and find the order of the polynomial

    2. Relevant equations



    3. The attempt at a solution

    b) I don't think it is diffiicult to find the linear combination; c*(a_+ a_+)=c*2*(K/2)^(1/2); Not sure how I would show that it does not normalized for the first excited state but would I start the problem in this manner? a* [itex]
    \varphi_1(x)
    [/itex]=0, to find [itex]
    \varphi_1(x)
    [/itex] and plug
    [tex]

    \int_{-\infty}^{\infty} |\varphi_1(x)|^2 dx=1

    [/tex] ; How would I not know that x[itex]
    \varphi_0(x)
    [/itex] is not normalizable?

    c) [tex]
    a_{+}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x+\left(\frac{1}{2k/(2m)^(1/2)}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0
    [/tex]
    Not sure what they are asking for.
     
    Last edited: Aug 24, 2009
  2. jcsd
  3. Aug 24, 2009 #2
    The problem asks you to analyze the ground state of the harmonic oscillator wavefunction without explicitly getting into special functions. (The form of the ground state wavefunction is rather simple.)

    You have been given the action of the raising and lowering operators in the position basis, which turn out to be differential equations of constraint on the ground state. But while I understand that

    [tex]a_{-}|\psi_{0}\rangle = 0[/tex]

    as this is the defining equation for the ground state, I don't see how [itex]a_{+}|\psi_{0}\rangle = 0[/itex]. In fact, the left hand side equals (give or take a constant factor) the ket [itex]|\psi_{1}\rangle[/itex].

    [Edit: Can you clean up your latex a bit, esp for your attempted solution of part (c).]
     
  4. Aug 24, 2009 #3

    gabbagabbahey

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    Your problem statement for b) is not very clear. What exactly are you being asked to express as a linear combination of raising and lowering operators? The [itex]X[/itex] operator? The [itex]P[/itex] operator? Something else entirely?

    Also, you are using an incorrect expression for the raising and lowering operators in the x-basis....why is there a [itex](2m)^{1/2}[/itex] in your expression?
     
  5. Aug 24, 2009 #4
    [tex]

    a_{-}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2/(\hbar/(2m))}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0

    [/tex]

    [tex]

    a_{+}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x+\left(\frac{1}{2(\hbar(2m)/)}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0

    [/tex]

    plancks constant, K parameter constant and m should all be within the operator equations; I am still trying to get acquainted with using latex; planck constant should be the numerator btw; These are the equations I am told to work with; Now that I have phi_0 from part a, I need to used the expression [tex]\varphi_0[/tex]*x=0 to show that the expression does not normalized; How would I know that the expression does not normalized?

    Should I express x in the terms operator a_-
     
    Last edited: Aug 24, 2009
  6. Aug 24, 2009 #5

    gabbagabbahey

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    It seems to me that you are supposed to first express the operator [itex]X[/itex] in terms of the raising and lowering operators, then use that to show that [itex]X|\varphi_0\rangle\propto|\varphi_1\rangle[/itex]
     
  7. Aug 24, 2009 #6
    Would I integrate x using the operators;


    [tex]


    a_{+}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x+\left(\frac{1}{2(\hbar(2m)/)}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0


    [/tex] and
    [tex]


    a_{-}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2/(\hbar/(2m))}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0


    [/tex]; Sorry for all the questions; I am working all by myself and usually don't join study groups;
     
  8. Aug 24, 2009 #7

    gabbagabbahey

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    What do you mean 'integrate x'?

    You have an expression for the raising an lowering operators,

    [tex]a_{+}=\left(\frac{k}{2}\right)^{1/2}x+\frac{\hbar}{(2m)^{1/2}}\frac{d}{dx}[/tex]

    [tex]a_{-}=\left(\frac{k}{2}\right)^{1/2}x-\frac{\hbar}{(2m)^{1/2}}\frac{d}{dx}[/tex]

    So what is the operator [itex]x[/itex] in terms of those two raising and lowering operators?

    What does that make [itex]x\varphi_0[/itex]?
     
  9. Aug 24, 2009 #8
    [tex]a_{+}+\frac{\hbar}{(2m)^{1/2}}\frac{d}{dx}[/tex]=[tex]\left(\frac{k}{2}\right)^{1/2}x[/tex]

    [tex]a_{-}-\frac{\hbar}{(2m)^{1/2}}\frac{d}{dx}[/tex]=[tex]\left(\frac{k}{2}\right)^{1/2}x[/tex],

    [itex]x\varphi_0[/itex] would be from part a of the problem correct?

    I ended up with this expression; x=(2/k)^(1/2)*(a_-+a_+)
     
    Last edited: Aug 24, 2009
  10. Aug 24, 2009 #9

    gabbagabbahey

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    That's not what I meant.

    You have two equations, involving the operators 'x', 'd/dx' and the raising and lowering operators. You should be able to combine your two equations to eliminate 'd/dx' and leave you with a single equation that involves only 'x' and the raising and lowering operators. You can then solve that equation for 'x'.

    No, in part (a) you found [itex]\varphi_0(x)[/itex], but you don't need that result for part (b)
     
  11. Aug 24, 2009 #10
    did you see my recently edited post
     
  12. Aug 24, 2009 #11
    I get it now; x [tex]
    \varphi_0
    [/tex]x=[tex]
    \varphi_0
    [/tex](2/k)^(1/2)*(a_-+a_+) and I would integrate this expression to obtain a new expression for [tex]
    \varphi_0
    [/tex] I don't think this expression is normalizable because everytime I attempt to integrate it [tex]
    \varphi_0
    [/tex] cancels itself out
     
  13. Aug 24, 2009 #12

    gabbagabbahey

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    I did now.

    Yes.

    No. How would integrating the expression for the operator 'x' give you an expression for [itex]\varphi_0[/itex]?!:confused:

    Instead just multiply your expression for x by [itex]\varphi_0[/itex]:

    [tex]x\varphi_0=\left(\frac{2}{k}\right)^{1/2}\left(a_-+a_+\right)\varphi_0=\left(\frac{2}{k}\right)^{1/2}a_{-}\varphi_0+\left(\frac{2}{k}\right)^{1/2}a_{+}\varphi_0[/tex]

    But from the definition of the lowering operator, [itex]a_{-}\varphi_0=[/itex]____? And from thew definition of the raising operator, [itex]a_{+}\varphi_0=[/itex]____?
     
  14. Aug 25, 2009 #13
    I know from the definition for the lower operator , [itex]\varphi_0[/itex]*a_=0 and therefore x =(2/k)^(1/2) a_+. Now I would apply E_n=(.5+n)

    ground state E_0=.5*omega*[tex]\hbar[/tex]
    first excited state E_1=3/2*omega*[tex]\hbar[/tex]
     
  15. Aug 25, 2009 #14

    gabbagabbahey

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    Yes.

    No.

    x=(2/k)^(1/2) (a_+ + a_-)

    Just because the effect of acting a_- on a certain state produces a result of zero does not mean that a_- is zero.

    You can however say that [itex]x\varphi_0=\left(\frac{2}{k}\right)^{1/2}a_+ \varphi_0[/itex]

    Not quite, this is not about energy eigenvalues. This is about the effect of the raising operator on a given eigenstate....when the raising operator acts on an eigenstate, it transforms it into another eigenstate with a higher energy. By definition, [itex]\varphi_0[/itex] is the lowest energy eigenstate, so the eigenstate with an energy level of one quanta higher is by definition, [itex]\varphi_1[/itex].

    Therefor [itex]a_+\varphi_0=\varphi_1[/itex]....follow?

    So [itex]x\varphi_0=[/itex]____?
     
  16. Aug 25, 2009 #15


    [itex]x\varphi_0=\left(\frac{2}{k}\right)^{1/2}\varphi_1[/itex]
     
  17. Aug 25, 2009 #16

    gabbagabbahey

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    Right, which is a non-normalized version of the 1st excited state. ([itex]\varphi_1[/itex] is presumably normalized, so [itex]\left(\frac{2}{k}\right)^{1/2}\varphi_1[/itex] is presumably unnormalized)
     
  18. Aug 25, 2009 #17
    in part c) E_n=(n+1/2)*omega [tex]
    \hbar
    [/tex] should I transformed this expression to the gaussian expression?

    Would this equation be relevant: H(a*[tex]\varphi[/tex] )=(E-[tex]\hbar[/tex] *omega)(a*[tex]\varphi[/tex] )
     
  19. Aug 26, 2009 #18
    err... bump! just like in part b, do I need to show that x* [tex]
    \varphi_n
    [/tex]=(2/K)^(1/2)*[tex]
    \varphi_(n+1)
    [/tex]

    1 is supposed to be added to the subscript along with n
     
  20. Aug 26, 2009 #19

    gabbagabbahey

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    Not quite, since [itex]a_{-}\varphi_{n-1}[/itex] is only equal to zero if n=1.....what do you get when you apply your expression for the x-operator to [itex]\varphi_{n-1}[/itex]?....can you use that, along with the fact that in part (a) you found that [itex]\varphi_0(x)[/itex] was Gaussian, to argue that [itex]\varphi_n=P(x)e^{-cx^2}[/itex] for some polynomial [itex]P(x)[/itex]?
     
  21. Aug 26, 2009 #20
    x*[itex]\varphi_{n-1}[/itex]= (2/K)^(1/2)*[tex] \varphi_(n) [/tex],since
    [itex]
    \varphi_n=P(x)e^{-cx^2}
    [/itex], then


    x*[itex]\varphi_{n-1}[/itex]= (2/K)^(1/2)*[tex] \varphi_(n) [/tex]= x*[itex]\varphi_{n-1}[/itex]= (2/K)^(1/2)*[itex]
    \varphi_n=P(x)e^{-cx^2}
    [/itex], then





    x*[itex]\varphi_{n-1}[/itex]=(2/K)^(1/2)*[itex]
    \varphi_n=P(x)e^{-cx^2}
    [/itex]
     
    Last edited: Aug 26, 2009
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