Complex scalar field -- Quantum Field Theory -- Ladder operators

[binbagsss]

Homework Statement

STATEMENT
$\hat{H}=\int \frac{d^3k}{(2\pi)^2}w_k(\hat{a^+(k)}\hat{a(k)} + \hat{b^{+}(k)}\hat{b(k)})$
where $w_k=\sqrt{{k}.{k}+m^2}$

The only non vanishing commutation relations of the creation and annihilation operators are:
$[\alpha(k),\alpha^{+}(p)] =(2\pi)^3 \delta^3(k-p)=[b(k),b^{+}(p)]$
(I have dropped hats on the alpha here and have done for the rest of the problem)

QUESTION
By calculating an expression for $<\psi|H|\psi>$ where $|\psi>$ is a normalised eigenstate of the Hamiltonian, show tht the energy is non-negative?

EQUATIONS
see above.
ATTEMPT
To be honest I really have no idea where to start.
Many books I've seen define the vacuum state and then compute states and the eigenvalues from there using the ladder operators and the commutation relationships, so I really do not no where to get started.

I think I may need some explicit form of $|\psi>$ to work with - do I first of all write down some general form of the eigenstate with ladder operators and then I can proceed as usual using the commutator relationships? Not sure how to do this though?

Many thanks in advance.

Homework Equations

The Attempt at a Solution

 

[binbagsss]

STATEMENT

$\hat{H}=\int \frac{d^3k}{(2\pi)^2}w_k(\hat{a^+(k)}\hat{a(k)} + \hat{b^{+}(k)}\hat{b(k)})$
where $w_k=\sqrt{{k}.{k}+m^2}$

The only non vanishing commutation relations of the creation and annihilation operators are:
$[\alpha(k),\alpha^{+}(p)] =(2\pi)^3 \delta^3(k-p)=[b(k),b^{+}(p)]$
(I have dropped hats on the alpha here and have done for the rest of the problem)

QUESTION

Assuming that for any state $<\psi|\psi>=0 \iff |\psi>=0$ (1) show that for all $k$, $a(k)$ and $b(k)$ annihilate the state $|0>$ which is defined as a normalized zero energy eigenstate.

EQUATIONS
see above.

ATTEMPT
To set $|\psi>=H|0>=0|0>$ and then use the (1) this gives $<\psi|\psi>=0$:

$<0|\int \frac{d^3k}{(2\pi)^3}w_k(a^{+}(k)a(k)+b^{+}(k)b(k)) \int \frac{d^3 p}{(2\pi)^3} w_p (a^{+}(p)a(p)+b^{+}(p)b(p)|0>$

I'm not sure whether this is the right thing to do, the double intergral looks a bit messy and I'm unsure what to do now...

Many thanks in advance.

 

[eys_physics]

In the derivation of your final formula you have use that $<\psi|=(|\psi>)^\dagger)$, or in other words
$<\psi|\psi>=<0|H^\dagger H|0>$. It is also a bit unclear if your creation operators correspond to fermions or bosons. In the first case you don't have commutators but anti-commutators.

 

[binbagsss]

In the derivation of your final formula you have use that $<\psi|=(|\psi>)^\dagger)$, or in other words
$<\psi|\psi>=<0|H^\dagger H|0>$. It is also a bit unclear if your creation operators correspond to fermions or bosons. In the first case you don't have commutators but anti-commutators.

The commutator relation was given and stated in the question, we can just take it and use it without needing to know what it corressponds to.

I don't understand what you are ssaying sorry.. $H^\dagger = H$, I have used this.

 

[binbagsss]

The commutator relation was given and stated in the question, we can just take it and use it without needing to know what it corressponds to.

I don't understand what you are ssaying sorry.. $H^\dagger = H$, I have used this.

$<\psi|=|\psi>^*=<0|H^{\dagger}$ is this not correct?