The last step of this Green's function proof is not clear

[Hill]

TL;DR Summary

In QFTftGA, Lancaster and Blundell show in a (1+1) dimensional spacetime that the function $G^+$ is a Green's function for the Schrodinger equation. I need the last step of this derivation to be clarified.

Here is the conclusion of the derivation in question:

where $\phi_n$ are eigenfunctions of the Hamiltonian.

I don't see how at the very end the $\sum ...$ becomes $\delta (x-y)$. What do I miss?

 

[Hill]

P.S. I thought of a way, but not sure in it:

1. For $t_x-t_y \neq 0$, the expression vanishes because of the $\delta (t_x-t_y)$ in front.
2. For $t_x-t_y = 0$,
$\sum_n \phi_n(x) \phi_n^*(y)=\sum_n \langle x| n\rangle \langle n|y \rangle = \langle x|y \rangle = \delta(x-y)$

I am not sure because the eigenstates $|n \rangle$ in the two brackets are at different times.

 

[Hill]

the eigenstates |n⟩ in the two brackets are at different times

Oh, sorry, they are not, in the case 2.
Solved.
Forget about it .

 

[haushofer]

For future reference: Hit the identity with a wavefunction f(y) and integrate over y. The integral gives a factor
</phi_n | f> giving the expansion of f(x) in the {/phi_n} basis. This is exactly what a delta function would do.

(On the phone so no fancy tex code)

Weird that the authors don't explain these steps in a book for "amateurs".

 

[Hill]

For future reference: Hit the identity with a wavefunction f(y) and integrate over y. The integral gives a factor
</phi_n | f> giving the expansion of f(x) in the {/phi_n} basis. This is exactly what a delta function would do.

I'm sorry, could you elaborate, please? I did not understand it in words.

 

[haushofer]

We start from the sum $\sum_n \phi_n (x) \phi^*_n (y)$ where the $\phi_n (x)$ form an orthonormal basis. Hit this expression with a function f(y) and integrate over y:

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy$

Now use that

$\int \phi^*_n (y) f(y)dy = <\phi_n | f> = c_n$ ("Fourier's trick)

such that

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = \sum_n c_n \phi_n (x)$

This is just the expansion of $f(x)$, so indeed

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = f(x)$

With other words:

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = \int \delta (x-y)f(y)dy$

So under the integral sign we have the equality

$\sum_n \phi_n (x) \phi^*_n (y) = \delta(x-y)$

:)

 

[Hill]

We start from the sum $\sum_n \phi_n (x) \phi^*_n (y)$ where the $\phi_n (x)$ form an orthonormal basis. Hit this expression with a function f(y) and integrate over y:

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy$

Now use that

$\int \phi^*_n (y) f(y)dy = <\phi_n | f> = c_n$ ("Fourier's trick)

such that

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = \sum_n c_n \phi_n (x)$

This is just the expansion of $f(x)$, so indeed

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = f(x)$

With other words:

$\int \sum_n \phi_n (x) \phi^*_n (y) f(y)dy = \int \delta (x-y)f(y)dy$

So under the integral sign we have the equality

$\sum_n \phi_n (x) \phi^*_n (y) = \delta(x-y)$

:)

Thank you!

 

[vanhees71]

TL;DR Summary: In QFTftGA, Lancaster and Blundell show in a (1+1) dimensional spacetime that the function $G^+$ is a Green's function for the Schrodinger equation. I need the last step of this derivation to be clarified.

Here is the conclusion of the derivation in question:

View attachment 336963

where $\phi_n$ are eigenfunctions of the Hamiltonian.

I don't see how at the very end the $\sum ...$ becomes $\delta (x-y)$. What do I miss?

Because of the factor $\delta(t_x-t_y)$ you can set $t_x=t_y$ in the rest of the expression on the right-hand side. This leads to
$$\sum_n \phi_n(x) \phi_n^*(y)=\sum_n \langle x|\phi_n \rangle \langle \phi_n |y\rangle = \langle x|y \rangle= \delta(x-y),$$
where I made use of the completeness of the energy eigenstates,
$$\sum_n |\phi_n \rangle \langle \phi_n=\hat{1},$$
and the normalization of the (generalized) position eigenvectors "to a $\delta$ distribution".