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8.2 Expectation value of \hat{{x}}^{2} and \hat{{p}}^{2} for the harmonic oscillator

As an example of all we have discussed let us look at the harmonic oscillator. Suppose we measure the average deviation from equilibrium for a harmonic oscillator in its ground state. This corresponds to measuring \hat{x}. Using

{ϕ}_{0}(x) ={ \left ({mω\over ℏπ} \right )}^{1∕4}\mathop{ exp}\nolimits \left (−{mω\over 2ℏ} {x}^{2}\right )
(8.17)

we find that

\langle x\rangle ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{\left ({mω\over ℏπ} \right )}^{1∕2}x\mathop{exp}\nolimits \left (−{mω\over ℏ} {x}^{2}\right )dx = 0.
(8.18)

Qn Why is it 0? Sinilarly, using \hat{p} = −iℏ {d\over dx} and

\hat{p}{\left ({mω\over ℏπ} \right )}^{1∕4}\mathop{ exp}\nolimits \left (−{mω\over 2ℏ} {x}^{2}\right ) = imωx{\left ({mω\over ℏπ} \right )}^{1∕4}\mathop{ exp}\nolimits \left (−{mω\over 2ℏ} {x}^{2}\right ),
(8.19)

we find

\langle p\rangle = 0.
(8.20)

More challenging are the expectation values of {x}^{2} and {p}^{2}. Let me look at the first one first:

\begin{eqnarray} \langle {x}^{2}\rangle & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{\left ({mω\over ℏπ} \right )}^{1∕2}{x}^{2}\mathop{ exp}\nolimits \left (−{mω\over ℏ} {x}^{2}\right )dx%& \\ & =& \left ( {ℏ\over mω}\right ){π}^{−1∕2}{\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{y}^{2}\mathop{ exp}\nolimits \left (−{y}^{2}\right )dy %& \\ & =& \left ( {ℏ\over mω}\right ){1\over 2}. %&(8.21) \\ \end{eqnarray}

Now for \hat{{p}}^{2},

\hat{{p}}^{2}\mathop{ exp}\nolimits \left (−{mω\over 2ℏ} {x}^{2}\right ) = [−{(mωx)}^{2} + ℏmω]\mathop{exp}\nolimits \left (−{mω\over 2ℏ} {x}^{2}\right ).
(8.22)

Thus,

\begin{eqnarray} \langle \hat{{p}}^{2}\rangle & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{\left ({mω\over ℏπ} \right )}^{1∕2}[−{(mωx)}^{2} + ℏmω]\mathop{exp}\nolimits \left (−{mω\over ℏ} {x}^{2}\right )dx%& \\ & =& ℏmω{π}^{−1∕2}{\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}[1 − {y}^{2}]\mathop{exp}\nolimits \left (−{y}^{2}\right )dy %& \\ & =& ℏmω{1\over 2}. %&(8.23) \\ \end{eqnarray}

This is actually a form of the uncertainity relation, and shows that

ΔxΔp ≥ {1\over 2}ℏ!
(8.24)

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