Comprehensive Study Notes

We derived the eigenfunctions of the linear-momentum and angular-momentum operators. We now ask: What are the eigenfunctions of the position operator $\hat{x}$ ?

The operator $\hat{x}$ is multiplication by $x$. Denoting the position eigenfunctions by $g_{a}(x)$, we write

$
\begin{equation}
x g{a}(x)=a g{a}(x) \tag{7.77}
\end{equation}
$

where $a$ symbolizes the possible eigenvalues. It follows that

$
\begin{equation}
(x-a) g_{a}(x)=0 \tag{7.78}
\end{equation}
$

We conclude from (7.78) that

$
\begin{equation}
g_{a}(x)=0 \quad \text { for } x \neq a \tag{7.79}
\end{equation}
$

Moreover, since an eigenfunction that is zero everywhere is unacceptable, we have

$
\begin{equation}
g_{a}(x) \neq 0 \quad \text { for } x=a \tag{7.80}
\end{equation}
$

These conclusions make sense. If the state function is an eigenfunction of $\hat{x}$ with eigenvalue $a, \Psi=g_{a}(x)$, we know (Section 7.6) that a measurement of $x$ is certain to give the value $a$. This can be true only if the probability density $|\Psi|^{2}$ is zero for $x \neq a$, in agreement with (7.79).

Before considering further properties of $g_{a}(x)$, we define the Heaviside step function $H(x)$ by (see Fig. 7.4)

$
\begin{array}{ll}
H(x)=1 & \text { for } x>0 \
H(x)=0 & \text { for } x<0 \tag{7.81}\
H(x)=\frac{1}{2} & \text { for } x=0
\end{array}
$

We next define the Dirac delta function $\delta(x)$ as the derivative of the Heaviside step function:

$
\begin{equation}
\delta(x) \equiv \frac{d H(x)}{d x} \tag{7.82}
\end{equation}
$

From (7.81) and (7.82), we have at once (see also Fig. 7.4)

$
\begin{equation}
\delta(x)=0 \quad \text { for } x \neq 0 \tag{7.83}
\end{equation}
$

FIGURE 7.4 The Heaviside step function.

Since $H(x)$ makes a sudden jump at $x=0$, its derivative is infinite at the origin:

$
\begin{equation}
\delta(x)=\infty \quad \text { for } x=0 \tag{7.84}
\end{equation}
$

We can generalize these equations slightly by setting $x=t-a$ and then changing the symbol $t$ to $x$. Equations (7.81) to (7.84) become

$
\begin{gather}
H(x-a)=1, \quad x>a \tag{7.85}\
H(x-a)=0, \quad x<a \tag{7.86}\
H(x-a)=\frac{1}{2}, \quad x=a \tag{7.87}\
\delta(x-a)=d H(x-a) / d x \tag{7.88}\
\delta(x-a)=0, \quad x \neq a \quad \text { and } \quad \delta(x-a)=\infty, \quad x=a \tag{7.89}
\end{gather}
$

Now consider the following integral:

$
\int_{-\infty}^{\infty} f(x) \delta(x-a) d x
$

We evaluate it using integration by parts:

$
\begin{gathered}
\int u d v=u v-\int v d u \
u=f(x), \quad d v=\delta(x-a) d x
\end{gathered}
$

Using (7.88), we have

$
\begin{gather}
d u=f^{\prime}(x) d x, \quad v=H(x-a) \
\int{-\infty}^{\infty} f(x) \delta(x-a) d x=\left.f(x) H(x-a)\right|{-\infty} ^{\infty}-\int{-\infty}^{\infty} H(x-a) f^{\prime}(x) d x \
\int{-\infty}^{\infty} f(x) \delta(x-a) d x=f(\infty)-\int_{-\infty}^{\infty} H(x-a) f^{\prime}(x) d x \tag{7.90}
\end{gather}
$

where (7.85) and (7.86) were used. Since $H(x-a)$ vanishes for $x<a$, (7.90) becomes

$
\begin{align}
& \int{-\infty}^{\infty} f(x) \delta(x-a) d x=f(\infty)-\int{a}^{\infty} H(x-a) f^{\prime}(x) d x \
&=f(\infty)-\int{a}^{\infty} f^{\prime}(x) d x=f(\infty)-\left.f(x)\right|{a} ^{\infty} \
& \int_{-\infty}^{\infty} f(x) \delta(x-a) d x=f(a) \tag{7.91}
\end{align}
$

Comparing (7.91) with the equation $\sum{k} c{k} \delta{i k}=c{i}$, we see that the Dirac delta function plays the same role in an integral that the Kronecker delta plays in a sum. The special case of (7.91) with $f(x)=1$ is

$
\int_{-\infty}^{\infty} \delta(x-a) d x=1
$

The properties (7.89) of the Dirac delta function agree with the properties (7.79) and (7.80) of the position eigenfunctions $g_{a}(x)$. We therefore tentatively set

$
\begin{equation}
g_{a}(x)=\delta(x-a) \tag{7.92}
\end{equation}
$

To verify (7.92), we now show it to be in accord with the Born postulate that $|\Psi(a, t)|^{2} d a$ is the probability of observing a value of $x$ between $a$ and $a+d a$. According to (7.76), this probability is given by

$
\begin{equation}
\left|\left\langle g{a}(x) \mid \Psi(x, t)\right\rangle\right|^{2} d a=\left|\int{-\infty}^{\infty} g_{a}^{}(x) \Psi(x, t) d x\right|^{2} d a \tag{7.93}
\end{}
$

Using (7.92) and then (7.91), we have for (7.93)

$
\left|\int_{-\infty}^{\infty} \delta(x-a) \Psi(x, t) d x\right|^{2} d a=|\Psi(a, t)|^{2} d a
$

which completes the proof.
Since the quantity $a$ in $\delta(x-a)$ can have any real value, the eigenvalues of $\hat{x}$ form a continuum: $-\infty<a<\infty$. As usual for continuum eigenfunctions, $\delta(x-a)$ is not quadratically integrable (Prob. 7.43).

Summarizing, the eigenfunctions and eigenvalues of position are

$
\begin{equation}
\hat{x} \delta(x-a)=a \delta(x-a) \tag{7.94}
\end{equation}
$

where $a$ is any real number.
The delta function is badly behaved, and consequently the manipulations we performed are lacking in rigor and would make a mathematician shudder. However, one can formulate things rigorously by considering the delta function to be the limiting case of a function that becomes successively more peaked at the origin (Fig. 7.5). The delta function is not really a function but is what mathematicians call a distribution (see en.wikipedia .org/wiki/Dirac_delta_function).

FIGURE 7.5 Functions that approximate $\delta(x)$ with successively increasing accuracy. The area under each curve is 1 .