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--- continuous_basis_representations [2021/03/12 20:13] – admin
+++ continuous_basis_representations [2021/03/12 21:25] (current) – [2.vi.3 Connecting the Position and Momentum Representations] admin
@@ Line 21: / Line 21: @@
 & = \int_{-\infty}^{+\infty}\D k\, f(k)^*\bra{\chi_k},
 \end{align*}
-where we can think of $f^*(k)$ as the components of a "row vector" with a continuous index.
+where we can think of $f(k)^*$ as the components of a "row vector" with a continuous index.
 As another example, we can calculate an inner product in a particular basis via:
@@ Line 28: / Line 28: @@
 & = \int_{-\infty}^{+\infty} \D k\, \braket{\phi}{\chi_k} \braket{\chi_k}{\psi} \\
 & = \int_{-\infty}^{+\infty} \D k\, \braket{\chi_k}{\phi}^* \braket{\chi_k}{\psi} \\
-& = \int_{-\infty}^{+\infty} \D k\, g^*(k) f(k),
+& = \int_{-\infty}^{+\infty} \D k\, g(k)^* f(k),
 \end{align*}
 where $g(k)$ are the components of $\ket{\phi}$ in the $\ket{\chi_k}$ basis, i.e. $\ket{\phi} = \int_{-\infty}^{+\infty} \D k\, g(k) \ket{\chi_k}$.
@@ Line 41: / Line 41: @@
 where we think of $A_{kk'} = \sand{\chi_k}{\hat{A}}{\chi_{k'}}$ as the components of a "matrix" with a continuous infinity of rows and columns.
-x====== 2.vi.1 The Position Representation ======
+====== 2.vi.1 The Position Representation ======
 Just as in the discrete case, there are an infinite number of orthonormal bases that we could use.  However, for most calculations, it is convenient to use just two: the position basis and the momentum basis.
@@ Line 61: / Line 61: @@
 \[\hat{\vec{r}} \ket{\vec{r}} = \vec{r}\ket{\vec{r}},\]
 which is understood to mean
-\[\left ( \begin{array}{c} \hat{x}\ket{\vec{r}} \\ \hat{y}\ket{\vec{r}} \\ \hat{z}\ket{\vec{r}} \end{array}\right ) = \left ( \begin{array}{c} x\ket{\vec{r}} \\ \y\ket{\vec{r}} \\ z\ket{\vec{r}} \end{array}\right ).\]
+\[\left ( \begin{array}{c} \hat{x}\ket{\vec{r}} \\ \hat{y}\ket{\vec{r}} \\ \hat{z}\ket{\vec{r}} \end{array}\right ) = \left ( \begin{array}{c} x\ket{\vec{r}} \\ y\ket{\vec{r}} \\ z\ket{\vec{r}} \end{array}\right ).\]
 The orthonormality relations for the position basis are
@@ Line 76: / Line 76: @@
 \[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty}\D x\, \ket{x}\braket{x}{\psi} = \int_{-\infty}^{+\infty}\D x\, \psi(x)\ket{x},\]
 where the function $\psi(x)$ is known as the (position space) //**wavefunction**// in quantum mechanics.
-p
 Similarly, in three-dimensions we have
 \[\ket{\psi} = \hat{I}\ket{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{r}\, \ket{\vec{r}}\braket{\vec{r}}{\psi} = \int_{\mathbb{R}^3}\D^3 \vec{r}\, \psi(\vec{r})\ket{\vec{r}},\]
@@ Line 86: / Line 86: @@
 \[\hat{p}\ket{p} = p\ket{p}.\]
 In three-dimensions, we introduce the three momentum operators $\hat{p}_x$, $\hat{p}_y$, $\hat{p}_z$ and the vector of operators
-\[\hat{\vec{p}} = \left ( \begin{array}{c} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \right ),\]
+\[\hat{\vec{p}} = \left ( \begin{array}{c} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{array}\right ),\]
 so that we can write the defining eigenvalue equations as
 \[\hat{\vec{p}} \ket{\vec{p}} = \vec{p}\ket{\vec{p}},\]
 where
-\[\vec{p} = \left ( \begin{array}{c} p_x \\ p_y \\ p_z \right ).\]
+\[\vec{p} = \left ( \begin{array}{c} p_x \\ p_y \\ p_z \end{array} \right ).\]
 The momentum basis states satisfy the orthonormality and completeness relations:
@@ Line 100: / Line 100: @@
 \[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty}\D p\, \ket{p}\braket{p}{\psi} = \int_{-\infty}^{+\infty}\D p\, \Psi(p)\ket{p},\]
 or, in three dimensions,
-\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{p}\, \ket{\vec{p}}\braket{\vec{p}}{\psi} = \int_{\mathbb{R}^3}\D^3 \vec{p}\, \psi(\vec{p})\ket{\vec{p}},\]
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{p}\, \ket{\vec{p}}\braket{\vec{p}}{\Psi} = \int_{\mathbb{R}^3}\D^3 \vec{p}\, \psi(\vec{p})\ket{\vec{p}},\]
 where $\Psi(p) = \braket{p}{\psi}$ and $\Psi(\vec{p}) = \braket{\vec{p}}{\psi}$ are called the //**momentum space wavefunction**//.
@@ Line 115: / Line 115: @@
 ====== 2.vi.3 Connecting the Position and Momentum Representations ======
-The position and momentum bases are different bases for the same Hilbert space, so there must be a way of converting between the position and momentum bases.  The Broglie and Planck hypotheses provide the connection.  They say that the position wavefunction of a state of definite momentum is a plane wave with wave vector $\vec{k} = \vec{p}/\hbar$ and angular frequency $\omega = E/\hbar$, $\psi(x,t) \propto e^{ikx - \omega t}$.  Consider the situation at $t=0$ so that $\psi(x) = \psi(x,0) \propto e^{ikx}$.  By the de Broglie hypothesis we can write $\psi(x) = e^{ipx/\hbar}$, so a definite momentum state can be written in the position basis as
+The position and momentum bases are different bases for the same Hilbert space, so there must be a way of converting between the position and momentum bases.  The Broglie and Planck hypotheses provide the connection.  They say that the position wavefunction of a state of definite momentum is a plane wave with wave vector $\vec{k} = \vec{p}/\hbar$ and angular frequency $\omega = E/\hbar$, $\psi(x,t) \propto e^{ikx - \omega t}$.  Consider the situation at $t=0$ so that $\psi(x) = \psi(x,0) \propto e^{ikx}$.  By the de Broglie hypothesis we can write $\psi(x) \propto e^{ipx/\hbar}$, so a definite momentum state can be written in the position basis as
 \[\ket{p} = A\int_{-\infty}^{+\infty}\D x\,e^{ipx/\hbar}\ket{x}.\]
 In an in class activity, you will determine that the normalization constant is $A = \frac{1}{\sqrt{2\pi\hbar}}$, so
@@ 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 ======