- #1
jim burns
- 5
- 0
I want to calculate $$\langle x|XP|y \rangle$$ where X is the position operator and P the momentum operator, and the states are position eigenstates. But I get two different answers depending on if I insert a complete set of states.
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?
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?