- #1

Able

- 8

- 0

Hi all. Here is an n-particle hamltonian.

[tex]H=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}+U(x)\Bigr)a(x)+\int d^{3}xd^{3}yV(x-y)a^{\dagger}(x)a^{\dagger}(y)a(x)a(y)[/tex]

Here is an n-particle state.

[itex]\Bigr|\psi,t\Bigl\rangle=\int d^{3}x_{1}...d^{3}x_{n}\psi(x_{1},...,x_{n};t)a^{\dagger}(x_{1})...a^{\dagger}(x_{n})\Bigr|0\Bigl\rangle[/itex]

I am wondering how this hamiltonian acts on this state. Let us first look at the kinetic term. Is it true to say that first the [itex]a(x)[/itex] in the hamiltonian searches through the state for it's counterpart [itex]a^{\dagger}(x)[/itex] and annihalates the particle at [itex]x[/itex]. Then the [itex]\nabla^{2}[/itex] acts on an [itex](n-1)[/itex]-particle state. Finally the [itex]a^{\dagger}(x)[/itex] restores the particle at [itex]x[/itex]? In the same way, the interaction hamiltonian acts on the state by removing two particles, operating with the interaction operator [itex]V[/itex] and finally restoring particles at the two points in question? If this is correct why do we think of it this way?

I am also worried about the dimension of the fields [itex]a(x)[/itex] and [itex]a^{\dagger}(x)[/itex]. I compared dimensions on the left and right hand sides of the hamiltonian and got something strange. What should I be expecting for the dimension of such a field?

I would also like a better understanding of the state above. I am used to thinking of states as a linear combination of chosen base states, the coefficients being the amplitudes for the state to be in one of the base states. For example,

[itex]\Bigl|\psi\Bigl\rangle=\underset{i}{\sum}\Bigl|i\Bigl\rangle\Bigr\langle i\Bigl|\psi\Bigl\rangle[/itex]So how am I to compare the two states? What is similar what is different? I can see how the sum and integral are related. Is the [itex]\psi(x_{1},...,x_{2};t)[/itex] just like the coefficients and the [itex]a^{\dagger}[/itex]'s acting on the vacuum state like the base states?

Cheers.

[tex]H=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}+U(x)\Bigr)a(x)+\int d^{3}xd^{3}yV(x-y)a^{\dagger}(x)a^{\dagger}(y)a(x)a(y)[/tex]

Here is an n-particle state.

[itex]\Bigr|\psi,t\Bigl\rangle=\int d^{3}x_{1}...d^{3}x_{n}\psi(x_{1},...,x_{n};t)a^{\dagger}(x_{1})...a^{\dagger}(x_{n})\Bigr|0\Bigl\rangle[/itex]

I am wondering how this hamiltonian acts on this state. Let us first look at the kinetic term. Is it true to say that first the [itex]a(x)[/itex] in the hamiltonian searches through the state for it's counterpart [itex]a^{\dagger}(x)[/itex] and annihalates the particle at [itex]x[/itex]. Then the [itex]\nabla^{2}[/itex] acts on an [itex](n-1)[/itex]-particle state. Finally the [itex]a^{\dagger}(x)[/itex] restores the particle at [itex]x[/itex]? In the same way, the interaction hamiltonian acts on the state by removing two particles, operating with the interaction operator [itex]V[/itex] and finally restoring particles at the two points in question? If this is correct why do we think of it this way?

I am also worried about the dimension of the fields [itex]a(x)[/itex] and [itex]a^{\dagger}(x)[/itex]. I compared dimensions on the left and right hand sides of the hamiltonian and got something strange. What should I be expecting for the dimension of such a field?

I would also like a better understanding of the state above. I am used to thinking of states as a linear combination of chosen base states, the coefficients being the amplitudes for the state to be in one of the base states. For example,

[itex]\Bigl|\psi\Bigl\rangle=\underset{i}{\sum}\Bigl|i\Bigl\rangle\Bigr\langle i\Bigl|\psi\Bigl\rangle[/itex]So how am I to compare the two states? What is similar what is different? I can see how the sum and integral are related. Is the [itex]\psi(x_{1},...,x_{2};t)[/itex] just like the coefficients and the [itex]a^{\dagger}[/itex]'s acting on the vacuum state like the base states?

Cheers.

Last edited: