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

Question

$\hat{H}=\int \frac{d^3k}{(2\pi)^2}w_k(\hat{a^+(k)}\hat{a(k)} + \hat{b^{+}(k)}\hat{b(k)})$ 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 $\langle\psi|H|\psi\rangle$ where $|\psi\rangle$ is a normalised eigenstate of the Hamiltonian, show tht the energy is non-negative?

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 $|\psi\rangle$ 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.

Answer 1

The calculation is actually trivial in this setting. The only thing to note is that $a^+$ and $a$ are adjoint operators of each other. That means that for any states $\psi,\phi$, you have $$\langle\psi|a^+(k)\phi\rangle = \langle a(k)\psi|\phi\rangle$$

Therefore you can easily write $$\begin{align} \langle\psi | H | \psi\rangle &= \int \frac{d^3k}{(2\pi)^2}w_k(\langle \psi | a^+(k)a(k)|\psi\rangle + \langle\psi|b^{+}(k)b(k)|\psi\rangle) \\ &= \int \frac{d^3k}{(2\pi)^2}w_k(|| a(k)|\psi\rangle||^2 + ||b(k)|\psi\rangle||^2) \end{align}$$ In this expression, $||\cdot||$ is always non-negative, because it is a norm of some vector in a Hilbert space. And $\omega_k$ is non-negative as well.

EDIT: Some clarification in response to comment:

• The Hamiltonian $H$ is the operator of the observable "Energy". So $\langle \psi | H | \psi \rangle$ is the expectation value of the energy in the state $\psi$. This is true for any normalized state, not just eigenstates of $H$.
• My proof did not require $\psi$ to be an eigenstate of $H$. But if it is, we would have $H|\psi\rangle=E|\psi\rangle$ for some number $E$.
• If you integrate something non-negative, the result will be non-negative as well. It doesn't matter what value the integration-varibales take ($k_{x,y,z}$ indeed run from negative to positive infinity). Only the integrand needs to be non-negative.
• If you have any operator $A$, the product $A^+A$ will indeed be a non-negative operator. But squaring $H$ as in $H^+H$ is not useful if you want to prove that $H$ itself is non-negative. Instead you need to find a different operator whoose square is $H$. In this case "$H=a^+a$" (up to the integral and $\omega$ prefactor).