1.15: Expectation values of k and \(\omega\)
- Page ID
- 50116
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The expectation value of k is obtained by integrating the wavefunction over all k. This must be performed in k-space
\[ \langle k\rangle = \frac{\frac{1}{2\pi}\int^{+\infty}_{-\infty}A(k)^{*}kA(k)dk}{\frac{1}{2\pi}\int^{+\infty}_{-\infty}A(k)^{*}A(k)dk} = \frac{\langle A|k|A\rangle}{\langle A|A\rangle} \nonumber \]
From the Inverse Fourier transform in k-space
\[ \psi(x)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}A(k)e^{ikx}dx \nonumber \]
note that
\[ -i\frac{d}{dx}\psi(x)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}kA(k)e^{ikx}dx \nonumber \]
Thus, we have the following Fourier transform pair:
\[ -i \frac{d}{dx}\psi(x) \Leftrightarrow kA(k) \nonumber \]
It follows that\(^{†}\)
\[ \langle k\rangle = \frac{\langle A\rangle|k|\langle A\rangle}{\langle A\rangle|\langle A\rangle} = \frac{\langle \psi | -i\frac{d}{dx}|\psi\rangle}{\langle \psi|\psi\rangle} \nonumber \]
Similarly, from the Inverse Fourier transform in the frequency domain
\[ \psi(t)=\frac{1}{2\pi}\int^{+\infty}_{-\infty} A(\omega)e^{-i\omega t} d\omega \nonumber \]
We can derive the Fourier transform pair:
\[ i \frac{d}{dt} \psi(t) \Leftrightarrow \omega A(\omega) \nonumber \]
It follows that
\[ \langle \omega \rangle = \frac{\langle A\rangle|\omega |\langle A\rangle}{\langle A\rangle|\langle A\rangle} = \frac{\langle \psi | i\frac{d}{dt}|\psi\rangle}{\langle \psi|\psi\rangle} \nonumber \]
We define two operators
\[ \hat{k} = -i \frac{d}{dx} \nonumber \]
and
\[ \hat{\omega} = i \frac{d}{dt} \nonumber \]
Operators only act on functions to the right. To signify this difference we mark them with a caret.
We could also define the (somewhat trivial) position operator
\[ \hat{x} = x \nonumber \]

