An interpretation of hamiltonians, states and fields.

[Able]

Hi all. Here is an n-particle hamltonian.

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)

Here is an n-particle state.

\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

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 a(x) in the hamiltonian searches through the state for it's counterpart a^{\dagger}(x) and annihalates the particle at x. Then the \nabla^{2} acts on an (n-1)-particle state. Finally the a^{\dagger}(x) restores the particle at x? In the same way, the interaction hamiltonian acts on the state by removing two particles, operating with the interaction operator V 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 a(x) and a^{\dagger}(x). 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,

\Bigl|\psi\Bigl\rangle=\underset{i}{\sum}\Bigl|i\Bigl\rangle\Bigr\langle i\Bigl|\psi\Bigl\rangleSo 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 \psi(x_{1},...,x_{2};t) just like the coefficients and the a^{\dagger}'s acting on the vacuum state like the base states?

Cheers.

 

[Able]

Erm, how do I make that code into something legible?

 

[kith]

Put it into [*tex] [*/tex] or [*itex] [*/itex] (without the *s)

 

[strangerep]

Hi all. Here is an n-particle hamltonian.

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)

Here is an n-particle state.

\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

I am wondering how this hamiltonian acts on this state.

First, do you understand how H acts on a very simple single-particle state?
What are your commutation relations? To check that you understand this much,
what is the result of the following?
<br /> H_0 a^\dagger(x) |0\rangle<br />
where H_0 is your H with U = V = 0.

And what is
<br /> H_U \; a^\dagger(x) |0\rangle ~~~~~~~ ?<br />
(where H_U is just the part of H containing U).

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 [psi] just like the coefficients and the a*'s acting on the vacuum state like the base states?

That's the general idea.

 

[Able]

Thanks for the reply strangerep.

Acting the hamiltonian on the single particle state, we get a(x)a^{\dagger}(x) hitting the vacuum state. Using the bosonic commutation relation \left[a(x),a^{\dagger}(x^{\prime})\right]=\delta(x-x^{\prime}) we can write a^{\dagger}(x)a(x)+1 instead. a(x) kills the vacuum state so we are left with

H_{0}a^{\dagger}(x)\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle

So from here I suppose we need the explicit form of the vacuum state and the field to do the differentiation and integration. Any ideas what they may look like? For a particular case even?

I cannot see how H_{U} would be any different

H_{U}a^{\dagger}(x)\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)U(x)\Bigl|0\Bigl\rangle

Able.

 

[strangerep]

Acting the hamiltonian on the single particle state, we get a(x)a^{\dagger}(x) hitting the vacuum state.

No. The x inside the Hamiltonian is a dummy variable of integration.
Choose a different symbol for one of the variables.

Using the bosonic commutation relation \left[a(x),a^{\dagger}(x^{\prime})\right]=\delta(x-x^{\prime}) we can write a^{\dagger}(x)a(x)+1 instead.

That doesn't follow -- but fix up the first mistake and it might become clearer.

H_{0}a^{\dagger}(x)\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle

You have a free variable x on the LHS but a dummy integration variable x on
the RHS. So something has gone wrong. Try doing it again, but taking more
care to distinguish between free and dummy variables.

 

[Able]

Ok, what about this

H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)a(x)a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle

H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\Bigl(a^{\dagger}(x_{1})a(x)+\delta^{3}(x-x_{1})\Bigr)\Bigl|0\Bigl\rangle

H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)a^{\dagger}(x_{1})a(x)\Bigl|0\Bigl\rangle+\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\delta^{3}(x-x_{1})\Bigl|0\Bigl\rangle

The first term in this is zero because the a(x) hits the vacuum state. Integrating the delta function gives

H_{0}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle=a^{\dagger}(x_{1})\Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle

Sorry about the inconsistency of factors.

 

[strangerep]

[...] Integrating the delta function gives

H_{0} a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle = a^\dagger (x_{1}) \Bigl(-\nabla^{2}\Bigr)\Bigl|0\Bigl\rangle

No. The derivative of the delta was wrt x, which is a parameter inside the delta,
so this must be handled correctly.

As a simpler warm-up exercise, do you know how the derivative of a delta function
<br /> \partial_x \; \delta(x - y)<br />
is defined (and handled in practical situations)? E.g., can you evaluate the following?
<br /> \int dx\; f(x) \; \partial_x \delta(x - y)<br />

 

[Able]

Well I didn't off hand but I did some research. It seems to only be defined in conjunction with another function inside an integral. Then you use integration by parts to put derivatives on the other function.

\int g(x)\delta^{\prime}(x-y)dx=-g^{\prime}(y)

\int g(x)\delta^{\prime\prime}(x-y)dx=g^{\prime\prime}(y)

So what I want is this

\int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\delta^{3}(x-x_{1})\Bigl|0\Bigl\rangle=-\frac{\hbar^{2}}{2m}\nabla^{2}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle

 

[strangerep]

Well I didn't off hand but I did some research. It seems to only be defined in conjunction with another function inside an integral. Then you use integration by parts to put derivatives on the other function.
[...]

Good. BTW, it would be less error-prone if you include a subscript on the derivatives
to indicate which variable they're for. E.g.,
<br /> \nabla^{2}_x ~~~\mbox{or}~~~ \nabla^{2}_{x_1} ~~~~ \mbox{(etc)}<br />
to make your next equation's rhs more explicit.

So what I want is this
<br /> \int d^{3}xa^{\dagger}(x)\Bigl(-\frac{\hbar^{2}}{2m}\nabla^{2}\Bigr)\delta^{3}(x-x_{1})\Bigl|0\Bigl\rangle<br /> ~=~ -\frac{\hbar^{2}}{2m}\nabla^{2}a^{\dagger}(x_{1})\Bigl|0\Bigl\rangle

Now, what equation does your free field satisfy in the absence of interactions?