Question 3.3: Find the eigenfunctions and eigenvalues of the position oper...

Let g_y(x) be the eigenfunction and y the eigenvalue:

\widehat{x} g_{y}(x)=x g_{y}(x)=y g_{y}(x)          (3.37)

Here y is a fixed number (for any given eigenfunction), but x is a continuous variable. What function of x has the property that multiplying it by x is the same as multiplying it by the constant y? Obviously it’s got to be zero, except at the one point x=y ; in fact, it is nothing but the Dirac delta function:

g_{y}(x)=A\delta (x-y)

This time the eigenvalue has to be real; the eigenfunctions are not square integrable, but again they admit Dirac orthonormality:

\int_{-\infty }^{\infty }{g^{*}_{\acute{y} }(x) g_{y} (x) } dx=\left|A\right|^{2} \int_{-\infty } ^{\infty } {\delta (x-\acute{y})\delta (x-y)}dx= \left|A \right |^{2}\delta (y-\acute{y})        (3.38)

If we pick A=1 , so

g_{y}(x)=\delta (x-y)       (3.39)

then

\left\langle g_{\acute{y}}|g_{y}\right\rangle \delta (y-\acute{y})       (3.40)

These eigenfunctions are also complete:

f(x)=\int_{-\infty }^{\infty }c(y){g_{y}(x)} dy=\int_{-\infty }^{\infty }{c(y)\delta (x-y)}dy        (3.41)

with

c(y)=f(y)       (3.42)

(trivial, in this case, but you can get it from Fourier’s trick if you insist).