[NOTES/QM-10002] Coordinate Representation

A Useful result 

The canonical commutation relation $[\widehat{x},\widehat{p}]=i\hbar$ can be used to show that if $x_0$ is an eigenvalue of $\widehat{x}$, $x_0+a$, $a\in{\mathbb R}$, is also an eigenvalue. This is most easily seen by making use of the
\begin{equation} \widehat{x} \exp(-ia\widehat{p}/\hbar) = \exp(-ia\widehat{p}/\hbar)(\widehat{x}+a) \label{EQ01}. \end{equation}
here $a$ is a real number. To see this note that $U(a)\equiv \exp(-ia\widehat{p}/\hbar)$ is a unitary operator. If $\ket{x_0}$ is an eigenvector of $\widehat{x}$ with eigenvalue $x_0$, $U(a)\ket{x_0}$ is an eigenvector of $\widehat{x}$ with eigenvalue $x_0+a$. Using \eqref{EQ01} we get \begin{eqnarray} \widehat{x} \Big(U(a)\ket{x_0}\Big) &=& \exp(-ia\widehat{p}/\hbar)(\widehat{x}+a)\ket{x_0}, \label{EQ02}\\ &=& (x_0+a) \Big(U(a)\ket{x_0}\Big) \label{EQ03}. \end{eqnarray} This shows that $(x_0+a)$ is an eigenvalue of $\widehat{x}$ with $U(a) \ket{x_0}$ as the eigenvector. Since $a$ is any real number, all real numbers are eigenvalues of the position operator $\widehat{x}$. A similar argument shows that all real values are allowed as eigenvalues of the momentum operator $\widehat{p}$.

Coordinate representation --- One dimension

We shall first consider a particle in one dimension. To set up the coordinate representation, we use the eigenvectors of position operator $\widehat{x}$ as o.n. basis. We have seen that the eigenvalues of $\widehat{x}$ are all real values in range $(-\infty, \infty)$. Let $x$ be one such eigenvalue and $\ket{x}$ be the corresponding eigenvector, {\it i.e.}, \begin{equation} \widehat{x} \ket{x} = x \ket{x}. \label{EQ04} \end{equation} The orthogonality property of the eigenvectors now assumes the form \begin{equation} \innerproduct{x^{\prime\prime}}{x^\prime}= \delta(x^{\prime\prime}-x^\prime) \label{EQ05}, \end{equation} and the completeness relation $\sum_n \ket{n}\bra{n} =\widehat{I}$ takes the form \begin{equation} \int_{-\infty}^\infty \ket{x}\bra{x}\, dx = \widehat{I}.\label{EQ06} \end{equation} where $\widehat{I}$ denotes the identity operator. Everywhere the sum $\sum_n$ over all eigenvalues is replaced by integration, $\int dx$, as the eigenvalues are now continuous. This choice of the eigenvectors \(\{\ket{x}\}\) as a basis leads to the coordinate representation or the position representation, also known as the \it Schr\"{o}dinger representation. Thus an expansion of abstract vector $\ket{\psi}$ in the basis $\big\{\ket{x}\big\}$ becomes \begin{equation} \ket{\psi} = \int \, dx \ket{x}\innerproduct{x}{\psi},\label{EQ07} \end{equation} and the abstract vector $\ket{\psi}$ is represented by the numbers $\innerproduct{x}{\psi}$, with $x$ having values in real numbers. Instead of arranging all the components of $\ket{\psi}$ in form of a column, we regard them as values of a function of $x$: \begin{equation} \innerproduct{x}{\psi} = \psi(x) \label{EQ08}. \end{equation}

• Due to the fact that the eigenvalues of $x$ are continuous, it is not meaningful to ask for probability that the position has a single value $x_0$; instead we must ask for probability position has a value in the specified range, such as $x$ and $x+ dx$.}

The third postulate tells us that this probability is given by \begin{equation} |\innerproduct{x}{\psi}|^2 dx = |\psi(x)|^2dx \label{EQ09}. \end{equation} Thus the function $\psi(x)$, to be called the {\bf wave function}, gives the probability density of position. The Parseval relation \begin{equation} \innerproduct{\psi}{\psi} = \int_{-\infty}^\infty |\psi(x)|^2\, dx \label{EQ10} \end{equation} ensures that, for normalized state vector \(\ket{\psi}\), the total probability will be one. Every operator $\widehat{T}$ is represented as a infinite dimensional matrix with continuous row and column indices $\big\{ ({\sf T})_{x,x^\prime}=\matrixelement{x}{\widehat{T}}{x^\prime} \big\}$. The action of operator $\widehat{T}$ on wave function $\psi(x)$ is then given by \begin{equation} \widehat{T} \psi(x) =\sum_{x^\prime} ({\sf T})_{x,x^\prime}\, \psi(x^\prime) =\int_{-\infty}^\infty \matrixelement{x}{\widehat{T}}{x^\prime}\, \psi(x^\prime)\, dx^\prime \label{EQ20}. \end{equation} The operator $\widehat{x}$ will be represented by an infinite dimensional diagonal matrix having rows and columns labelled by continuous indices $x^\prime, x^{\prime\prime}$ and the matrix elements of $x$ are \begin{equation} \matrixelement{x^\prime}{\widehat{x}}{x^{\prime\prime}} = x^\prime \delta(x^{\prime\prime}- x^\prime)\label{EQ11}. \end{equation} Using the canonical commutation relation, $[\widehat{x},\widehat{p}]=i\hbar$, the matrix elements of the momentum operator can be worked out and is given by \begin{equation} \matrixelement{x^\prime}{\widehat{p}}{x^{\prime\prime}} = -i\hbar\dd{x^\prime} \delta(x^{\prime\prime}- x^\prime)\label{EQ12}. \end{equation}

Action of position and momentum operators

The action of position and momentum operators is given as matrix multiplication by considering these operators as matrices with continuous row and column indices: \begin{eqnarray} \widehat{x} \psi(x) &=&\int_{-\infty}^\infty \matrixelement{x}{\widehat{x}}{x^\prime}\psi(x^\prime)\, dx^\prime \label{EQ13}\\ &=& \int_{-\infty}^\infty x\delta(x-x^\prime)\psi(x^\prime)\, dx^\prime \label{EQ18}\\ &=& x \psi(x) \end{eqnarray} and \begin{eqnarray} \widehat{p} \psi(x) &=& \int_{-\infty}^{\infty} \matrixelement{x}{\widehat{p}}{x^\prime}\psi(x^\prime)\, dx^\prime \label{EQ14}\\ &=& \int_{-\infty}^{\infty} -i\hbar\delta(x-x^\prime)\psi(x^\prime)\, dx^\prime \label{EQ15}\\ &=& -i\hbar \dd[\psi(x)]{x} \label{EQ17}. \end{eqnarray} Thus, in the position representation we have $\widehat{p} \to -i\hbar\dd{x}$. An operator corresponding to a dynamical variable can be obtained by making a replacement \begin{equation} \widehat{x} \to x, \qquad \qquad \widehat{p} \to -i\hbar \dd{x}.\label{EQ21} \end{equation}

Hamiltonian in coordinate representation

The most important dynamical variable for a system is the Hamiltonian \begin{equation} H =\frac{p^2}{2m} + V(x) \label{EQ22}. \end{equation} and the corresponding operator is \begin{equation} \widehat{H} = \frac{\widehat{p}}{2m} + V(\widehat{x}) = -\frac{\hbar^2}{2m}\DD{x} + V(x).\label{EQ23} \end{equation}

Wave function as position probability density

Let $\ket{\psi}$ denote the state vector of a particle and $\psi(x)=\innerproduct{x}{\psi}$ be the corresponding wave function. We now come to the physical interpretation of the coordinate space wave function $\psi(x)$. Note that $\psi(x)$ is the coefficient of $\ket{x}$ in the expansion of the state vector $\ket{\psi}$: \begin{equation} \ket{\psi} = \int \ket{x} \innerproduct{x}{\psi} \, dx = \int \psi(x) \ket{x} \, dx \label{EQ29}. \end{equation} The third postulate tells us that the expansion coefficient will give the probability amplitude for position and the the absolute square $|\psi(x)|^2$ gives the probability density for position. Thus the probability of finding position in the range $x, x+dx$ is $|\psi(x)|^2 dx$. The integral $\int_a^b |\psi(x)|^2\,dx $ is the probability that the particle will be found in the interval $(a,b)$.

Several degrees of freedom

Generalization to a particle in three dimensions is straightforward. The basis vectors $\ket{\vec{r}}$ in this case are simultaneous eigenvectors of position operators $\widehat{x}, \widehat{y},\widehat{z}$: \begin{equation} \widehat{x} \ket{\vec{r}} = x \ket{\vec{r}}, \qquad \widehat{y} \ket{\vec{r}} = y \ket{\vec{r}}, \qquad \widehat{z} \ket{\vec{r}} = z \ket{\vec{r}}.\label{EQ24} \end{equation} The orthogonality and completeness relations assume the form \begin{equation} \innerproduct{\vec{r}^{\,\prime\prime}}{\vec{r}^{\,\prime}} = \delta (\vec{r}^{\,\prime\prime}-\vec{r}^{\,\prime}), \qquad \int d^3\vec{r}\, \ket{\vec{r}}\bra{\vec{r}} = \widehat{I}. \label{EQ25s} \end{equation} Expansion of an arbitrary $\ket{\psi}$ in the basis $\ket{\vec{r}}$ assumes the form \begin{equation} \ket{\psi} = \int \ket{\vec{r}} \, \innerproduct{\vec{r}}{\psi} d^3r. \label{EQ26} \end{equation} We call the function $\innerproduct{\vec{r}}{\psi}$ the wave function and also denote it by $\psi(\vec{r})$. The absolute square of wave function gives the probability density; $|\psi(\vec{r})|^2 dV$ is the probability density for the particle to be a small volume $dV$ at position \(\vec{r}\). The corresponding probability for a a particle to be in a finite volume V is obtained integrating over the volume $V$ and is given by \begin{equation} \iiint_V |\innerproduct{\psi}{\vec{r}}|^2\, d^3r = \iiint_V |\psi(\vec{r})|^2\, d^3r \label{EQ27}. \end{equation} The action of position operators $\widehat{\vec{r}}$ is to multiply by $\vec{r}$ and that of momentum operators is given by $\widehat{\vec{p}} \to -i\hbar \nabla$. Thus \begin{equation} \widehat{\vec{r}} \,\psi(\vec{r}) =\vec{r}\, \psi(\vec{r}), \qquad \widehat{\vec{p}}\, \psi(\vec{r}) =-i\hbar \nabla \psi(\vec{r}) \label{EQ28}. \end{equation}