Multi-Particle QM Homework: Hamiltonian & Vacuum State

[binbagsss]

Homework Statement

Question attached :

Just number 1 please

Homework Equations

Hamiltonian is given by eq ? In the attachment

$\phi(x_1,x_2,...x_n,t)= \int dx_1...dx_n \phi(x_1,x_2...x_n,t) a^+(x_1)...a^(x_n) |0>$

Where $|0>$ is the vacuum state and $a(x) |0>=0$

The Attempt at a Solution

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I can follow the solution and it make sense wanting to move the derivative from delta to the wave function, however, I thought the gradient operator would act on everything to the right, and so you would use a product rule here..so we would get a term where it acts in the delta and the w.f multiplies it as given, as well as and a term where it acts on the wave function and the delta just multiplies it

Solution continued here :

Many thanks

 

[TSny]

...
as well as and a term where it acts on the wave function and the delta just multiplies it

The $\nabla^2$ operator involves derivatives with respect to the coordinates associated with $\vec x$ and not with respect to the coordinates of $\vec x_1$ or $\vec x_2$. Since $\psi$ is a function of $\vec x_1$ and $\vec x_2$, but not of $\vec x$, the operator $\nabla^2$ acting on the wavefunction gives zero.

When $\nabla^2$ acts on $\delta^{(3)}(\vec x - \vec x_1)$ you can show $\nabla^2 \delta^{(3)}(\vec x - \vec x_1) = \nabla_1^2 \delta^{(3)}(\vec x - \vec x_1)$ where $\nabla_1^2$ acts on the coordinates of $\vec x_1$. This allows you to integrate by parts twice (with respect to the coordinates of $\vec x_1$) to make $\nabla_1^2$ act on the wavefunction.

 

[binbagsss]

The $\nabla^2$ operator involves derivatives with respect to the coordinates associated with $\vec x$ and not with respect to the coordinates of $\vec x_1$ or $\vec x_2$. Since $\psi$ is a function of $\vec x_1$ and $\vec x_2$, but not of $\vec x$, the operator $\nabla^2$ acting on the wavefunction gives zero.

.

Omg, thank you !

 

[binbagsss]

Omg, thank you !

The $\nabla^2$ operator involves derivatives with respect to the coordinates associated with $\vec x$ and not with respect to the coordinates of $\vec x_1$ or $\vec x_2$. Since $\psi$ is a function of $\vec x_1$ and $\vec x_2$, but not of $\vec x$, the operator $\nabla^2$ acting on the wavefunction gives zero.

When $\nabla^2$ acts on $\delta^{(3)}(\vec x - \vec x_1)$ you can show $\nabla^2 \delta^{(3)}(\vec x - \vec x_1) = \nabla_1^2 \delta^{(3)}(\vec x - \vec x_1)$ where $\nabla_1^2$ acts on the coordinates of $\vec x_1$. This allows you to integrate by parts twice (with respect to the coordinates of $\vec x_1$) to make $\nabla_1^2$ act on the wavefunction.

Hi ok I've done this but for the x integration to remove the dx and delta, there are no terms to replace x by $x_1$. ... ?

 

[TSny]

Hi ok I've done this but for the x integration to remove the dx and delta, there are no terms to replace x by $x_1$. ... ?

There's the creation operator at the beginning of the integrand which depends on $\vec x$.

This $\vec x$ will get replaced by $\vec x_1$ when integrating over $d^3x$ due to the delta function $\delta^{(3)}(\vec x-\vec x_1)$.

 

[binbagsss]

There's the creation operator at the beginning of the integrand which depends on $\vec x$.
View attachment 235792
This $\vec x$ will get replaced by $\vec x_1$ when integrating over $d^3x$ due to the delta function $\delta^{(3)}(\vec x-\vec x_1)$.

A-ha I must be blind.