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


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?


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.


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).
  • $\begingroup$ For OP: This is more or less a direct adaptation of the proof of the positivity of the quantum mechanical harmonic oscillator. $\endgroup$ – Cameron Williams Apr 21 '17 at 11:54
  • $\begingroup$ also does the rest of the argument follow, to break it down very simply apologies, integrating a positive quantity yields a positive quantity if integrated over a positive region? but $k_x$ y, z etc I thought run from minus infinity to infinity. Also I see the question specifically mentions the fact the states are normalised, but your proof is independent of this fact since $ |AB| = |A||B| $ ? $\endgroup$ – yourlazyphysicist Mar 20 '18 at 18:43
  • $\begingroup$ Thank you for your reply Simon. I'm afraid my qm is quite poor so just to go back to basics the energy is the eigenvalue of the eigenstate, assuming no explicit time dependence which there isn't here: $ H | \phi> = E \phi >$ how then from $<\phi | H | \phi > $ is the RHS above the energy that we are looking at? from this I can construct $<\phi|H^+H|\phi> =|E|^2 <\phi|\phi> = |E|^2$ where I made have use of normalisation and (where the + represents the adjoint ) but not $<\phi | H | \phi > $. $\endgroup$ – yourlazyphysicist Mar 20 '18 at 18:46
  • $\begingroup$ I could probably talk in two different arbitrary states via (assuming these steps are valid, please let me know if they are not) : $<\phi | H^+ H | \phi > = <\phi | H^+ | H \phi > =<\phi | H | \phi' > $ where $H+H| \phi'> = E' |\phi' > $ ( and I'd also need to modify the eigenvalue on the right hand side, but all absorbed as a constant, positive again due to I have 'H^2' , and then use the fact that they are arbitrary to have the argument hold for the energy of any states ? $\endgroup$ – yourlazyphysicist Mar 20 '18 at 18:55
  • $\begingroup$ many thanks for your comments, they have helped. In the question we are told that the state is an eigenstate of the Hamiltonian. I see above you referred to E as 'some number' - but this would also be interpreted as the energy? If you're asked to compute the energy, I tthought you would use this eigenfunction equation, rather than it's expected value? Also, since the question specifies that the state is an eigenfunction of the Hamiltonian, I believe that there must be a proof that uses this, roughly how would this go? $\endgroup$ – yourlazyphysicist May 24 '18 at 14:53

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