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Maximum position expectation value for 1D harmonic oscillator

  1. Jun 18, 2015 #1
    Hey, I'm stuck halfway through the solution it seems. I could use some tips on how to continue.

    1. The problem statement, all variables and given/known data

    I have to determine a linear combination of the states [itex]|0\rangle, |1\rangle[/itex], of a one dimensional harmonic oscillator, so that the expectation value [itex]\langle x \rangle [/itex] is a maximum.

    2. Relevant equations
    [itex]x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)[/itex]

    3. The attempt at a solution
    I set [itex]|\alpha \rangle\equiv c_1|1\rangle + c_2|0\rangle[/itex]
    Then I calculate [itex] \langle \alpha | x | \alpha \rangle = \sqrt{\frac{\hbar}{2m\omega}} (c_1^*\langle1| + c_2^*\langle 0|)(a+a^\dagger)(c_1|1\rangle + c_2|0\rangle)=\sqrt{\frac{\hbar}{2m\omega}}[c_1^*(c_1+c_2)+c_2^*c_2] = \sqrt{\frac{\hbar}{2m\omega}}[1+c_1^*c_2][/itex]
    I get the last equation because of the normalization condition:
    [itex] c_1^*c_1+c_2^*c_2=1[/itex]
    This is where I don't know how to continue. Is this even the right approach?
    (I'm not allowed to use the wave functions.)

    Anything is appreciated.

    Kind regards
    Alex
     
  2. jcsd
  3. Jun 18, 2015 #2

    vela

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    You should recheck your calculation of ##\langle \alpha \lvert x \rvert \alpha \rangle##. The ##c_i^*c_i## terms shouldn't be there.
     
  4. Jun 18, 2015 #3

    blue_leaf77

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    As for the maximization problem, the Lagrange multiplier method may be helpful.
     
  5. Jun 18, 2015 #4
    Thanks for the fast reply.
    I've corrected my calculation, is this right?:
    First I calculate [itex](a+a^\dagger)|\alpha\rangle [/itex]
    [itex]a(c_1|1\rangle+c_2|0\rangle)=c_1|0\rangle+0c_2|0\rangle=c_1|0\rangle[/itex]
    and
    [itex]a^\dagger(c_1|1\rangle+c_2|0\rangle)=c_1\sqrt{2}|2\rangle+c_2|1\rangle[/itex]

    Then because [itex] |\alpha\rangle^\dagger = c_1^*\langle1|+c_2^*\langle0|[/itex]
    [itex]\langle\alpha|(a+a^\dagger)|\alpha\rangle=c_1^*\langle1|(c_1\sqrt{2}|2\rangle+c_2|1\rangle+c_1|0\rangle)+c_2^*\langle0|(c_1\sqrt{2}|2\rangle+c_2|1\rangle+c_1|0\rangle)= c_1^*c_2+c_2^*c_1[/itex]
    Do I use the lagrange multiplier method here? Or how do I continue?
     
  6. Jun 18, 2015 #5

    blue_leaf77

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    The Lagrange multiplier is used for finding the maximum value of <x>.
     
  7. Jun 18, 2015 #6
    So It's like a function of 4 variables [itex]f(c_1,c_1^*, c_2,c_2^*)=c_1^*c_2+c_2^*c_1[/itex] which I have to maximize under the condition [itex] c_1^*c_1+c_2^*c_2=1 [/itex]
    Or is there an easier solution?
     
  8. Jun 18, 2015 #7
    Hello! I Have the same exercise in my homework. Using the Lagrange multiplier I get ##c_{1}=c_{2}=1/\sqrt{2}##. Is that right?
     
  9. Jun 18, 2015 #8

    blue_leaf77

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    Yes, it is.
    Whether it's true or not, you can do some kind of examination. For example, you can write ##\langle \hat{x} \rangle \propto |c_1||c_2| \cos{(\phi_1-\phi_2)}## where the ##\phi##'s are the angle of each coefficients, since we want to maximize our expectation value, it's necessary that ##\phi_1=\phi_2##.
     
    Last edited: Jun 18, 2015
  10. Jun 19, 2015 #9
    Thanks for advise, blue_leaf77. I have made my calculation new.
    I get 4 different sets of values.
    ##c_{1}=c_{2} = i/\sqrt{2}##
    ##c_{1}=c_{2} = -i/\sqrt{2}##
    ##c_{1}=i/\sqrt{2}; c_{2} = -i/\sqrt{2}##
    ##c_{1}=-i/\sqrt{2}; c_{2} = i/\sqrt{2}##

    Using your advise how to examine solutions only the first and the second solutions are of interest. I can get this result on another way just inserting values in ##f## and finding a maximum. My question is: how i decide which set is right (the first or the second)?
     
  11. Jun 19, 2015 #10

    blue_leaf77

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    Well as long as ##\phi_1 = \phi_2## is satisfied, any ##\phi_1## is possible.
    In the original question posted by the OP, he is asked to calculate the maximum of ##\langle \hat{x} \rangle##. In your case, I don't know what you are asked to do in your own homework.
     
  12. Jun 19, 2015 #11
    I have to calculate the maximum of ##\langle \hat{x} \rangle## too
     
  13. Jun 19, 2015 #12

    blue_leaf77

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    You have this freedom.
     
  14. Jun 19, 2015 #13

    vela

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    Your first solution corresponds to the state
    $$\lvert \alpha_1 \rangle = i\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right) = e^{i\frac {\pi}{2}}\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right),$$ and your second solution to
    $$\lvert \alpha_2 \rangle = -i\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right) = e^{-i\frac {\pi}{2}}\left(\frac {1}{\sqrt{2}} \lvert 0 \rangle + \frac{1}{\sqrt{2}} \lvert 1 \rangle\right).$$ The two states differ only by an overall phase factor, so…
     
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