An interpretation of hamiltonians, states and fields.

1. Jun 22, 2011

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,

$\Bigl|\psi\Bigl\rangle=\underset{i}{\sum}\Bigl|i\Bigl\rangle\Bigr\langle i\Bigl|\psi\Bigl\rangle$

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 $\psi(x_{1},....,x_{2};t)$ just like the coefficients and the $a^{\dagger}$'s acting on the vacuum state like the base states?

Cheers.

Last edited: Jun 22, 2011
2. Jun 22, 2011

Able

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

3. Jun 22, 2011

kith

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

4. Jun 23, 2011

strangerep

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?
$$H_0 a^\dagger(x) |0\rangle$$
where H_0 is your H with U = V = 0.

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

That's the general idea.

5. Jun 23, 2011

Able

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.

Last edited: Jun 23, 2011
6. Jun 23, 2011

strangerep

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

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

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.

7. Jun 25, 2011

Able

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

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

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

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.

Last edited: Jun 25, 2011
8. Jun 25, 2011

strangerep

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
$$\partial_x \; \delta(x - y)$$
is defined (and handled in practical situations)? E.g., can you evaluate the following?
$$\int dx\; f(x) \; \partial_x \delta(x - y)$$

Last edited: Jun 25, 2011
9. Jun 25, 2011

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

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

Last edited: Jun 25, 2011
10. Jun 25, 2011

strangerep

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.,
$$\nabla^{2}_x ~~~\mbox{or}~~~ \nabla^{2}_{x_1} ~~~~ \mbox{(etc)}$$
to make your next equation's rhs more explicit.

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