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--- continuous_basis_representations [2021/03/09 00:17] – created admin
+++ continuous_basis_representations [2021/03/12 21:25] (current) – [2.vi.3 Connecting the Position and Momentum Representations] admin
@@ Line 11: / Line 11: @@
 \[\hat{I} = \int_{-\infty}^{+\infty} \D k \, \proj{\chi_k},\]
 and we can use this to expand any ket $\ket{\psi}$ in the $\ket{\chi_k}$ basis as
-\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty} \D k\,\ket{\chi_k}\braket{\chi_k}{\psi} = \int_{-\infty}^{+\infty}\D \k\, f(k)\ket{\chi_k},\]
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty} \D k\,\ket{\chi_k}\braket{\chi_k}{\psi} = \int_{-\infty}^{+\infty}\D k\, f(k)\ket{\chi_k},\]
 where $f(k) = \braket{\chi_k}{\psi}$ are the components of $\ket{\psi}$ in the $\ket{\chi_k}$ basis, with the main difference to what we have seen before being that $f(k)$ is now a function of a continuous variable rather than a discrete set of components.  We can think of it as a "column vector" with a continuous index.
@@ Line 19: / Line 19: @@
 & = \int_{-\infty}^{+\infty} \D k\,\braket{\psi}{\chi_k}\bra{\chi_k} \\
 & = \int_{-\infty}^{+\infty} \D k\,\braket{\chi_k}{\psi}^*\bra{\chi_k} \\
-& = \int_{-\infty}^{+\infty}\D \k\, f(k)^*\bra{\chi_k},
+& = \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 44: / Line 44: @@
 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.
+To construct the position basis, we first introduce the position operator $\hat{x}$ and define the basis states $\ket{x}$ to be the eigenstates of $\hat{x}$, i.e. $\hat{x}\ket{x} = x\ket{x}$.  The vector $\ket{x}$ is to be understood as representing the state of a quantum particle in one-dimension that has a definite position $x$.
+In three-dimensions, we have operators $\hat{x}$, $\hat{y}$, $\hat{z}$ representing the position coordinate in three perpendicular directions.  We then construct the position basis $\ket{\vec{r}}$, where $\vec{r}$ is the three dimensional position vector
+\[\vec{r} = \left ( \begin{array}{c} x \\ y \\ z \end{array}\right ).\]
+The defining properties of the position ket $\ket{\vec{r}}$ is that it is an eigenvector of all three coordinate operators
+\begin{align*}
+\hat{x} \ket{\vec{r}} & = x\ket{\vec{r}}, & \hat{y} \ket{\vec{r}} & = y\ket{\vec{r}}, & \hat{z} \ket{\vec{r}} & = z\ket{\vec{r}}.
+\end{align*}
+To make the notation a bit simpler, we introduce the somewhat bizarre concept of a three-dimensional column vector of operators, as follows.
+\[\hat{\vec{r}} = \left ( \begin{array}{c} \hat{x} \\ \hat{y} \\ \hat{z} \end{array}\right ).\]
+A ket $\ket{\psi}$ is understood to act like a scalar with respect to vectors of operators, i.e.
+\[\hat{\vec{r}}\ket{\psi} = \left ( \begin{array}{c} \hat{x}\ket{\psi} \\ \hat{y}\ket{\psi} \\ \hat{z}\ket{\psi} \end{array}\right ).\]
+This allows us to write the defining eigenvalue equations for $\ket{\vec{r}}$ as
+\[\hat{\vec{r}} \ket{\vec{r}} = \vec{r}\ket{\vec{r}},\]
+which is understood to mean
+\[\left ( \begin{array}{c} \hat{x}\ket{\vec{r}} \\ \hat{y}\ket{\vec{r}} \\ \hat{z}\ket{\vec{r}} \end{array}\right ) = \left ( \begin{array}{c} x\ket{\vec{r}} \\ y\ket{\vec{r}} \\ z\ket{\vec{r}} \end{array}\right ).\]
+The orthonormality relations for the position basis are
+\begin{align*}
+\braket{x}{x'} & = \delta(x-x'), & \braket{\vec{r}}{\vec{r}'} & = \delta(\vec{r} - \vec{r}'),
+\end{align*}
+and the completeness relations (decompositions of the identity) are
+\begin{align*}
+\int_{-\infty}^{+\infty} \D x\, \proj{x} & = \hat{I}, & \int_{\mathbb{R}^3} \D^3 \vec{r}\, \proj{\vec{r}} & = \hat{I}.
+\end{align*}
+For the uninitiated, $\int_{\mathbb{R}^3} \D^3 \vec{r} = \int_{-\infty}^{+\infty}\D x\,\int_{-\infty}^{+\infty}\D y\,\int_{-\infty}^{+\infty}\D z$ is just notation for a three-dimensional integral over all space.
+These relations allow us to do the Dirac notaty in the position representation.  For example, in one dimension
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty}\D x\, \ket{x}\braket{x}{\psi} = \int_{-\infty}^{+\infty}\D x\, \psi(x)\ket{x},\]
+where the function $\psi(x)$ is known as the (position space) //**wavefunction**// in quantum mechanics.
+Similarly, in three-dimensions we have
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{r}\, \ket{\vec{r}}\braket{\vec{r}}{\psi} = \int_{\mathbb{R}^3}\D^3 \vec{r}\, \psi(\vec{r})\ket{\vec{r}},\]
+where we now have a wavefunction $\psi(\vec{r})$ that is a function of all three coordinates.
+====== 2.vi.2 The Momentum Representation ======
+We can construct a basis of definite momentum states in the same way as we did for position.  In one-dimension, we introduce the momentum operator $\hat{p}$ and the definite momentum states $\ket{p}$ that satisfy the eigenvalue equation
+\[\hat{p}\ket{p} = p\ket{p}.\]
+In three-dimensions, we introduce the three momentum operators $\hat{p}_x$, $\hat{p}_y$, $\hat{p}_z$ and the vector of operators
+\[\hat{\vec{p}} = \left ( \begin{array}{c} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{array}\right ),\]
+so that we can write the defining eigenvalue equations as
+\[\hat{\vec{p}} \ket{\vec{p}} = \vec{p}\ket{\vec{p}},\]
+where
+\[\vec{p} = \left ( \begin{array}{c} p_x \\ p_y \\ p_z \end{array} \right ).\]
+The momentum basis states satisfy the orthonormality and completeness relations:
+\begin{align*}
+\braket{p}{p'} & = \delta(p-p'), & \braket{\vec{p}}{\vec{p}'} & = \delta(\vec{p} - \vec{p}') \\
+\int_{-\infty}^{+\infty} \D x\, \proj{p} & = \hat{I}, & \int_{\mathbb{R}^3} \D^3 \vec{p}\, \proj{\vec{p}} & = \hat{I},
+\end{align*}
+which allow us to find the components of any vector $\ket{\psi}$ in the momentum basis via
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{-\infty}^{+\infty}\D p\, \ket{p}\braket{p}{\psi} = \int_{-\infty}^{+\infty}\D p\, \Psi(p)\ket{p},\]
+or, in three dimensions,
+\[\ket{\psi} = \hat{I}\ket{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{p}\, \ket{\vec{p}}\braket{\vec{p}}{\Psi} = \int_{\mathbb{R}^3}\D^3 \vec{p}\, \psi(\vec{p})\ket{\vec{p}},\]
+where $\Psi(p) = \braket{p}{\psi}$ and $\Psi(\vec{p}) = \braket{\vec{p}}{\psi}$ are called the //**momentum space wavefunction**//.
+Note that the inner product can be computed either in the position basis
+\[\braket{\phi}{\psi} = \int_{-\infty}^{+\infty} \D x \, \braket{\phi}{x}\braket{x}{\psi} = \int_{-\infty}^{+\infty} \D x\, \phi^*(x)\psi(x),\]
+or the momentum basis
+\[\braket{\phi}{\psi} = \int_{-\infty}^{+\infty} \D p \, \braket{\phi}{p}\braket{p}{\psi} = \int_{-\infty}^{+\infty} \D p \, \Phi^*(p)\Psi(p),\]
+or in three-dimensions,
+\[\braket{\phi}{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{r} \, \braket{\phi}{\vec{r}}\braket{\vec{r}}{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{r} \, \phi^*(\vec{r})\psi(\vec{r}),\]
+or the momentum basis
+\[\braket{\phi}{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{p} \, \braket{\phi}{\vec{p}}\braket{\vec{p}}{\psi} = \int_{\mathbb{R}^3} \D^3 \vec{p}\, \Phi^*(\vec{p})\Psi(\vec{p}).\]
+Although the calculations you have to do in the position and momentum bases are different, they will always give the same result, since they are just computations of the same inner product in different bases.
+====== 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 $\psi(x) \propto e^{ipx/\hbar}$, 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
+\[\ket{p} = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty}\D x\,e^{ipx/\hbar}\ket{x}.\]
+From this, we can determine the inner product $\braket{x}{p}$.
+\begin{align*}
+\braket{x}{p} & = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D x'\, e^{ipx'} \braket{x}{x'} \\
+& = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{+\infty} \D x'\, e^{ipx'} \delta(x-x') \\
+& = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx}.
+\end{align*}
+This is so important that I will put it in a box.
+\[\boxed{\braket{x}{p} = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx}}\]
+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
+\[\ket{\vec{p}} = \frac{1}{\left ( 2\pi\hbar\right )^{3/2}}\int_{\mathbb{R}^3}\D^3 \vec{r} \, e^{i\vec{p}\cdot\vec{r}/\hbar} \ket{\vec{r}},\]
+yields
+\[\boxed{\braket{\vec{r}}{\vec{p}} = \frac{1}{\left ( 2\pi\hbar \right )^{3/2}} e^{i\vec{p}\cdot\vec{r}}.}\]
+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 ======