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Expectation values of x and x^2

  1. Apr 11, 2005 #1
    Given the wave function:
    [tex] \psi (x,t) = Ae^ {-\lambda \mid x\mid}e^ {-(i ) \omega t} [/tex]
    where A, [itex] \lambda [/itex], and [itex] \omega [/itex] are positive real constants

    I'm asked to find the expectation values of x and x^2.

    I know that the values are given by
    [tex] <x> = \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
    and
    [tex] <x^2> = \int_{-\infty}^{+\infty} (x^2)(A^2)e^ {-2\lambda \mid x\mid} dx [/tex]
    However, when calculated, I get <x> = <x^2> = 0. Since this would yield a standard deviation of zero, I'm thinking I've made a mistake (the reasoning being that the function does have some spread).

    Does this seem correct, or should I be getting a non-zero value for one of the expectation values?
     
  2. jcsd
  3. Apr 11, 2005 #2

    xanthym

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    SOLUTION HINTS:

    For <x>:

    [tex] 1: \ \ \ \ <x> \ \ = \ \ \int_{-\infty}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ [/tex]

    [tex] 2: \ \ \ \ \ \ \ = \ \ \int_{-\infty}^{0} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ [/tex]

    [tex] 3: \ \ \ \ \ \ \ = \ \ \color{red}\int_{0}^{+\infty} \color{black} \color{red}(-x)\color{black}(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} x(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ \color{red}\mathbf{(0)} [/tex]


    For <x2>:

    [tex] 4: \ \ \ \ <x^{2}> \ \ = \ \ \int_{-\infty}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ [/tex]

    [tex] 5: \ \ \ \ \ \ \ = \ \ \int_{-\infty}^{0} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ [/tex]

    [tex] 6: \ \ \ \ \ \ \ = \ \ \color{red}\int_{0}^{+\infty}\color{black} \color{red}(x^{2})\color{black}(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ + \ \ \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \mid x\mid} \, dx \ \ = \ \ \ [/tex]

    [tex] 7: \ \ \ \ \ \ \ = \ \ \color{red} (2) \cdot \color{black} \int_{0}^{+\infty} (x^{2})(A^2)e^ {-2\lambda \color{red} (x) \color{black} } \, dx \ \ \ \ = \ \ {\color{red}\mathbf{\displaystyle \left( \frac {A^{2}} {2\lambda^{3}}\right )}} [/tex]




    ~~
     
    Last edited: Apr 11, 2005
  4. Apr 11, 2005 #3
    Thanks. That's the justification I was looking for.

    Just to clearify the rule, let me see if I can generalize it...

    given:

    [tex] \int_{-\infty}^{+\infty} f(x) dx [/tex]

    Then, the integral can be re-written as:

    [tex] 2 \int_{0}^{+\infty} f(x) dx [/tex]

    if [itex] \int_{-\infty}^{0} f(x) dx = \int_{0}^{+\infty} f(x) dx [/itex]

    Is this correct?
     
    Last edited: Apr 11, 2005
  5. Apr 11, 2005 #4

    Galileo

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    It holds whenever the the integrand is even wtr 0. So f(x)=f(-x).

    If f(x)=-f(-x) the function is odd and the integral will be zero.
     
  6. Apr 11, 2005 #5

    dextercioby

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    Yeah,that modulus in the exponential surely means a lot.The exponential of real argument in neither odd,nor even,but that modulus changes things.

    Daniel.
     
  7. Apr 11, 2005 #6
    Thanks for the help. I should be set then.
     
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