- #1
T-7
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I wonder if someone could examine my argument for the following problem.
Using the relation
[tex]\widehat{x}^{2} = \frac{\hbar}{2m\omega}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )[/tex]
and properties of the ladder operators, determine the expectation value [tex]<\widehat{x}^{2}>[/tex] for the ground state of the simple harmonic well.
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )u_{0}.dx[/tex]
I then argue
[tex]
\left[ \widehat{A},\widehat{A}^{+}\right] = \widehat{A}\widehat{A}^{+} - \widehat{A}^{+}\widehat{A} = 1
[/tex]
[tex]
\Rightarrow \widehat{A}\widehat{A}^{+} = 1 + \widehat{A}^{+}\widehat{A}
\Rightarrow \widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2\widehat{A}^{+}\widehat{A} + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2n + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1
[/tex]
I am not quite clear if it is right that [tex]\widehat{A}^{+}\widehat{A}[/tex] is equal to n (the eigenfunction number), which is zero here. Could someone comment on that, and on whether or not my treatment above is ok?
Well, if this is true, then
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}( \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1)u_{0}.dx[/tex]
and by exploiting the orthonormal properties, I argue that the first two integrals are zero (you have one eigenfunction multiplied by another and integrated over infinity), but the third integral is 1, and then
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}[/tex]
Cheers!
Homework Statement
Using the relation
[tex]\widehat{x}^{2} = \frac{\hbar}{2m\omega}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )[/tex]
and properties of the ladder operators, determine the expectation value [tex]<\widehat{x}^{2}>[/tex] for the ground state of the simple harmonic well.
The Attempt at a Solution
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )u_{0}.dx[/tex]
I then argue
[tex]
\left[ \widehat{A},\widehat{A}^{+}\right] = \widehat{A}\widehat{A}^{+} - \widehat{A}^{+}\widehat{A} = 1
[/tex]
[tex]
\Rightarrow \widehat{A}\widehat{A}^{+} = 1 + \widehat{A}^{+}\widehat{A}
\Rightarrow \widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2\widehat{A}^{+}\widehat{A} + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2n + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1
[/tex]
I am not quite clear if it is right that [tex]\widehat{A}^{+}\widehat{A}[/tex] is equal to n (the eigenfunction number), which is zero here. Could someone comment on that, and on whether or not my treatment above is ok?
Well, if this is true, then
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}( \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1)u_{0}.dx[/tex]
and by exploiting the orthonormal properties, I argue that the first two integrals are zero (you have one eigenfunction multiplied by another and integrated over infinity), but the third integral is 1, and then
[tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}[/tex]
Cheers!