1

For a real scalar field (which is a bosonic field) we have these commutation relations :

$$ \left[\phi(x,t),\phi(y,t)\right]=0 \qquad \qquad \left[\phi(x,t),\pi(y,t)\right]=\delta(x-y).\tag{1}$$

But in condensed matter theory we usually write:

$$\left[\phi(x,t),\phi^\dagger (y,t)\right]=\delta(x-y) \tag{2} $$

if $\phi$ be a bosonic field.

My question is that if $\phi$ is a real field so it's conjugate must be itself. so :

$$\left[\phi(x,t),\phi^\dagger (y,t)\right]=\left[\phi(x,t),\phi(y,t)\right]=0.\tag{3}$$

What's wrong?

Qmechanic
  • 201,751
Hossein
  • 1,397

2 Answers2

3

The point is that the equation of motion of the fields is different if referring to temporal derivatives. In relativistic field theory, it is a second-order one and you need two initial conditions i.e. $\pi$ and $\pi$ to solve it. Quantizing, and interpreting the Fourier coefficients of the initial conditions as creation and annihilation operators, $$\phi(0,x) = \sum_k \frac{a_k}{\sqrt{2E_k}} e^{ikx} + \frac{a^*_k}{\sqrt{2E_k}} e^{-ikx}$$ $$\pi(0,x) = -i\sum_k \frac{a_k}{\sqrt{2E_k}} e^{ikx} - \frac{a^*_k}{\sqrt{2E_k}} e^{-ikx}$$ these quantized initial conditions turn out to have the standard fixed-time type of CCR as consequence of $$[a_k,a_h^*]= \delta_{kh}I\:, \quad [a_k,a_h]= 0 = [a^*_k,a_h^*]\tag{1}$$

In classical field theory, the equation is the Schroedinger one which, as it just involves the first derivative in $t$, needs only one initial condition. Moreover the field is complex. Also in this case, when interpreting the Fourier coefficients of the initial condition as creation and annihilation operators $$\phi(0,x) = \sum_k a_k e^{ikx}$$ $$\phi(0,x)^* = \sum_k a_k^* e^{-ikx}$$ you find the other type of CCR, again as a consequence of (1).

Valter Moretti
  • 71,578
1

Comments to the question (v3):

  1. Eqs. (1) are part of the CCR for a scalar field, such as, e.g., a real or complex Klein-Gordon field, a Schrödinger field, etc.

  2. Eq. (2) refers to the Schrödinger field, which is a complex field, see e.g. this Phys.SE post. A real Schroedinger field does naively not make sense since e.g. the expected kinetic term $\propto \phi\dot{\phi}$ would become a total time derivative.

Community
  • 1
Qmechanic
  • 201,751