Differences

This shows you the differences between two versions of the page.

--- continuous_basis_representations [2021/03/12 20:24] – [2.vi.3 Connecting the Position and Momentum Representations] admin
+++ continuous_basis_representations [2021/03/12 21:25] (current) – [2.vi.3 Connecting the Position and Momentum Representations] admin
@@ Line 126: / Line 126: @@
 \end{align*}
 This is so important that I will put it in a box.
-\[\boxed{\braket{x}{p} & = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx}}\]
+\[\boxed{\braket{x}{p} = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx}}\]
-To generalize this to three dimensions, not that the general form of a plane wave at time $t=0$ in three dimensions is $\psi(x) \propto e^{i\vec{k}\cdot\vec{r}} = e^{i\vec{p}\cdot\vec{r}/\hbar}$.  The normalization constant is $\left ( 2\pi\hbar\right )^{-3/2}$ and then the decomposition
+To generalize this to three dimensions, note that the general form of a plane wave at time $t=0$ in three dimensions is $\psi(x) \propto e^{i\vec{k}\cdot\vec{r}} = e^{i\vec{p}\cdot\vec{r}/\hbar}$.  The normalization constant is $\left ( 2\pi\hbar\right )^{-3/2}$ and then the decomposition
-\[\ket{\vec{p}} = \frac{1}{\left ( 2\pi\hbar\right )^{3/2}}\int_{\mathbb{R}^3}\D^3 \vec{r}  e^{i\vec{p}\cdot\vec{r}/\hbar} \ket{\vec{r}},\]
+\[\ket{\vec{p}} = \frac{1}{\left ( 2\pi\hbar\right )^{3/2}}\int_{\mathbb{R}^3}\D^3 \vec{r} \, e^{i\vec{p}\cdot\vec{r}/\hbar} \ket{\vec{r}},\]
 yields
-\[\boxed{\braket{\vec{r}}{\vec{p}}} = \frac{1}{\left ( 2\pi\hbar \right )^{3/2} e^{i\vec{p}\cdot\vec{r}}.}\]
+\[\boxed{\braket{\vec{r}}{\vec{p}} = \frac{1}{\left ( 2\pi\hbar \right )^{3/2}} e^{i\vec{p}\cdot\vec{r}}.}\]
 These inner products allow us to convert between the two different basis representations via
+\begin{align*}
+  \psi(x) & = \braket{x}{\psi}  \\
+  & = \sand{x}{\hat{I}}{\psi} \\
+  & = \int_{-\infty}^{+\infty} \D p\, \braket{x}{p}\braket{p}{\psi} \\
+  & = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D p\, e^{ipx/\hbar} \Psi(p).
+\end{align*}
+In other words, the position space wavefunction is the inverse Fourier transform of the momentum space wavefunction.
+\[\boxed{\psi(x) =\frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D p\, e^{ipx/\hbar} \Psi(p)}.\]
+In three dimensions, we will have
+\[\boxed{\psi(\vec{r}) =\frac{1}{\left ( 2\pi\hbar \right )^{3/2}} \int_{\mathbb{R}^3} \D^3 \vec{p}\, e^{i\vec{p}\cdot \vec{r}/\hbar} \Psi(\vec{p})}.\]
+Similarly, the momentum space wavefunction is given by
+\begin{align*}
+  \Psi(p) & = \braket{p}{\psi}  \\
+  & = \sand{p}{\hat{I}}{\psi} \\
+  & = \int_{-\infty}^{+\infty} \D x\, \braket{p}{x}\braket{x}{\psi} \\
+  & = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D x\, e^{-ipx/\hbar} \psi(x),
+\end{align*}
+which is the Fourier transform of the position space wavefunction.
+\[\boxed{\Psi(p) =\frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D x\, e^{-ipx/\hbar} \psi(x)},\]
+and in three dimensions this will be
+\[\boxed{\Psi(\vec{p}) =\frac{1}{\left ( 2\pi\hbar \right )^{3/2}} \int_{\mathbb{R}^3} \D^3 \vec{r}\, e^{-i\vec{p}\cdot \vec{r}/\hbar} \psi(\vec{r})}.\]
+====== 2.vi.4 The Momentum Operator in the Position Basis ======
+We know that, in the momentum basis $\hat{p} \ket{p} = p \ket{p}$, so the matrix elements of the momentum operator in the momentum basis are $\sand{p}{\hat{p}}{p'} = p \delta(p-p')$.  This allows us to compute the action of $\hat{p}$ on any ket $\ket{\psi}$ by working in the momentum basis.  If $\ket{\psi'} = \hat{p} \ket{\psi}$ then
+\begin{align*}
+\Psi'(p) & = \braket{p}{\psi'} \\
+& = \sand{p}{\hat{p}}{\psi} \\
+& = \sand{p}{\hat{p}\hat{I}}{\psi} \\
+& = \int_{-\infty}^{+\infty} \D p' \, \sand{p}{\hat{p}}{p'}\braket{p'}{\psi} \\
+& = \int_{-\infty}^{+\infty} \D p' \, p\delta(p-p')\Psi(p') \\
+& = p\Psi(p).
+\end{align*}
+In other words, we just multiply the momentum space wavefunction by $p$.
+However, what if we want to know how $\hat{p}$ acts on the position space wavefunction $\psi(x)$.  Recall, that the operator $\hat{\frac{\D}{\D x}}$ is defined through its action in the position basis via
+\[\hat{\frac{\D}{\D x}}\ket{\psi} = \int_{-\infty}^{+\infty} \D x\, \frac{\D \psi}{\D x} \ket{x}\]
+We can start by noticing that
+\[\frac{\D}{\D x} \left ( e^{ipx/\hbar} \right ) = \frac{ip}{\hbar}e^{ipx/\hbar},\]
+so that
+\[-i\hbar \frac{\D}{\D x} \left ( e^{ipx/\hbar} \right ) = p e^{ipx/\hbar}.\]
+Therefore, since we want $\hat{p}\ket{p} = p\ket{p}$ and we know that
+\[\ket{p} = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D p\,e^{ipx/\hbar} \ket{x},\]
+we have
+\[\hat{p}\ket{p} = -\i\hbar \hat{\frac{\D}{\D x}} \ket{p}.\]
+However, since $\ket{p}$ is a complete orthonormal basis, this implies that
+\[\boxed{\hat{p}} = -i\hbar \hat{\frac{\D}{\D x}},\]
+which means that, if $\ket{\psi'} = \hat{p}\ket{\psi}$ then
+\begin{align}
+\psi'(x) &  = \sand{x}{\hat{p}}{\psi} \\
+& = -i\hbar \sand{x}{\hat{\frac{\D}{\D x}}}{\psi} \\
+& = -i\hbar \int_{-\infty}^{+\infty} \D x'\, \frac{\D \psi}{\D x} \braket{x}{x'}  \\
+& = -i\hbar \int_{-\infty}^{+\infty} \D x'\, \frac{\D \psi}{\D x} \delta(x-x') \\
+& = -i\hbar \frac{\D \psi}{\D x},
+\end{align}
+i.e. we act with -i\hbar \frac{\D}{\D x} on the position space wavefunction $\psi(x)$.
+Note that it is common to use the sloppy notation
+\[\hat{p} \psi(x) = -i\hbar \frac{\D \psi}{\D x},\]
+which really means
+\[\sand{x}{\hat{p}}{\psi} = -i\hbar \frac{\D \braket{x}{\psi}}{\D x},\]
+but this causes no confusion in situations where we are //only// working in the position basis, which is often the case.
+In three dimensions, we will have the generalization
+\[\boxed{\hat{\vec{p}} = -i\hbar \hat{\vec{\nabla}}},\]
+where
+\[\vec{\nabla} = \left ( \begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array}\right ).\]
+====== 2.vi. 5 The Hamiltonian Operator in the Position Basis ======