I have some trouble understanding how one can, in the context of QFT, diagonalize a Hamiltonian $H$ by the introduction of ladder operators $a$ and $a^\dagger$ (I have trouble understanding how one is supposed to obtain these operators exactly).

As far as I understand, "diagonalizing" a Hamiltonian means finding ladder operators $a_p$ and $a_p^\dagger$ that obey the canonical commutation relation $$[a_p,a_k^\dagger]\propto \delta(k-p)\quad\text{and}\quad [a_k,a_p]=[a^\dagger_k, a^\dagger_p]=0$$ and rewrite $H$ in terms of these ladder operators such that $$H(a,a^\dagger)a^\dagger\vert0\rangle\propto a^\dagger\vert0\rangle\quad \text{and}\quad H(a,a^\dagger)a\vert n\rangle\propto a\vert n\rangle. $$

Lets assume that we want to diagonalize the Hamiltonian of the complex scalar field $\phi$ (which we obtained from the Lagragian $\mathcal{L}$), that is $$H=\int d^3x\left(\pi^\dagger \pi +\nabla\phi^\dagger\nabla\phi+m^2\phi^\dagger\phi\right).$$ This Hamiltonian is a function of $\phi$ and $\phi^\dagger$ (since one can express $\pi^\dagger=\dot\phi^\dagger$ and $\pi=\dot\phi$ in terms of those two). So what I'm now looking for are $a(\phi,\pi^\dagger)$, $a^\dagger(\phi,\pi^\dagger)$, $b(\phi,\pi^\dagger)$ and $b^\dagger(\phi,\pi^\dagger)$ (two sets of operators since $\phi$ and $\phi^\dagger$ are independent of each other).

Now most (all that I've seen so far) books/lecture notes just skip to assuming that we have $$\begin{align*} a_p&=\int d^3 x e^{ipx} (\omega_p \phi(x)+ i\pi^\dagger(x))\quad a_p^\dagger=\int d^3 x e^{ipx} (\omega_p \phi^\dagger(x)- i\pi(x))\\ b_p&=\int d^3 x e^{ipx} (\omega_p \phi^\dagger(x)+ i\pi(x))\quad b_p^\dagger=\int d^3 x e^{ipx} (\omega_p \phi(x)- i\pi^\dagger(x)) \end{align*}$$ and then show that they satisfy what we want. But how do we get to these opperators? I've found this SE post, where a method is proposed, but kind of fail to apply it to this example. If I have understood it correctly I should just assume that we have

$$a = \alpha \phi + \beta \pi^\dagger,\quad a^\dagger = \alpha^* \phi^\dagger + \beta^*\pi,$$ but what about the second pair of ladder operators and how to distinguish them form the first one? I'm really not sure if this works here...

TL;DR: I'd like to know how one can find ladder operators that diagonalize a given Hamiltonian $H(\phi,\pi)$ concretely.


2 Answers 2


The decomposition of the a and b that you refer to come from writing the most general solution to the (classical) equations of motion. This allows the identification of the physical degrees of freedom (a and b being essentially the Fourier coefficients of the solutions) which should be used for quantisation.

  • $\begingroup$ It is not a priori obvious that the a and b (the specific combination of the field and its momentum) will disgonalise H (and to do this it is not at all necessary that they obey the harmonic oscillator commutation relations) and it may be necessary in other theories to make further changes of variables to diagonalise H $\endgroup$
    – nox
    Commented Oct 5, 2019 at 17:00

You get these operators by noticing that the Hamiltonian looks like the oscillator Hamiltonian. The formula you wrote has a Fourier transform from the $x$ space to $p$ space. When you apply this transform to the Hamiltonian density, you get something that looks like the Oscillator Hamiltonian.

The formula also has $(x+ip)$ stuff, which is the standard formula for the creation and annihilation operators for the oscillator Hamiltonian $x^2+p^2$.

Look up R Shankar's Principle of Quantum Mechanics. Chapter : Time dependent Perturbation theory.


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