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Just as we have an abstract state vector $|\psi\rangle$ and its position representation $\psi(\vec{x}) = \langle \vec{x} | \psi \rangle$, how do we transform between a linear operator, say $H$, that acts on state vectors thus: $H |\psi\rangle$, and the differential operator form of the Hamiltonian that can act on the position representations of these states?

I have read that we have something like this

$$\langle \vec{x} | H | \vec{x}' \rangle = -\frac{h^2}{2m} \nabla^2 \delta(\vec{x} - \vec{x}') \,.$$

If this is so, could somebody explain how the following leap is made --- it concerns the propagator, which I'm defining by (with $\theta$ the Heaviside function):

$$K(\vec{x},t;\vec{x}',t') = \theta(t - t') \langle \vec{x} | U(t-t') |\vec{x}'\rangle $$

So I can't see why this line is true:

$$ \langle \vec{x} | H U(0) | \vec{x}' \rangle = H K(\vec{x},t';\vec{x}',t') \,,$$

where I assume that on the left hand side, the Hamiltonian is acting as an abstract linear operator, whilst on the right it takes its position representation, differential operator form. $U$ is the time evolution operator.

Thanks.

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1 Answer 1

up vote 1 down vote accepted

This simply follows from a resolution of the identity, written (formally) as $\int dy |y\rangle\langle y| = 1$:

$\langle x|HU(t)|x\rangle = \int dy\langle x|H|y\rangle\langle y|U(t)|x\rangle$, which after integrating out the delta function arising from the matrix element of $H$ gives you what you want.

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