- #1

jim burns

- 5

- 0

First way:

$$\langle x|XP|y \rangle=x\langle x|P|y \rangle=-ix \partial_x \delta(x-y)$$

Second way:

$$\langle x|XP|y \rangle=\int dz \langle x| z\rangle \langle z|XP|y \rangle=

\int dz \delta(x-z) [-iz\partial_z\delta(z-y)]

$$

Now to complete the second way, since there is $$\delta(x-z)$$, we can set z=x to get:

$$\langle x|XP|y \rangle=\int dz \langle x| z\rangle \langle z|XP|y \rangle=

\int dz \delta(x-z) [-iz\partial_z\delta(z-y)]=[-ix\partial_x\delta(x-y)]

$$

However, if we use the formula $$\int dz f(z)\partial_z \delta(z-y)=-\partial_y f(y)$$ we get:

$$\langle x|XP|y \rangle=\int dz \langle x| z\rangle \langle z|XP|y \rangle=

\int dz \delta(x-z) [-iz\partial_z\delta(z-y)]=\partial_y [+iy\delta(x-y)]=\\i\delta(x-y)+iy\partial_y\delta(x-y)=

i\delta(x-y)-ix\partial_x\delta(x-y)

$$

There seems to be an extra term $$i\delta(x-y)$$. Why doesn't this second way work?