# Hamiltonian of a quantum heat bath

I have seen the Hamiltonian for a heat bath written as:

$$H_B = \hbar \int_0^\infty \omega b(\omega)^\dagger b(\omega) d\omega$$

I was hoping to understand this equation better. This suggests that the heat bath is written as a sum of harmonic oscillators with raising/lowering operators $$b^\dagger(\omega)$$ and $$b(\omega)$$. I understand that is that this is just the continuous limit of the quantised electromagnetic field.

Please could someone explain the following:

1) Does the raising operator $$b^\dagger(\omega)$$ correspond directly to a mode in the heat bath with frequency $$\omega$$? ie does $$b^\dagger(\omega)b(\omega)$$ give us the number operator for the number of excitations of frequency $$\omega$$? or does the argument $$\omega$$ mean something else?

2) How does one alter this to change the temperature of the heat bath? My current interpretation is that this equation is for the vacuum/zero temperature case, so do we just add a Boltzmann factor inside the integral, to change the temperature or is there more subtlety required?

• 1) Yes, $b(\omega)$ corresponds to a mode of frequency $\omega$. 2) The Hamiltonian is always temperature independant. It goes into the Boltzmann factor $\exp(-\beta H_B)$, not the other way round. – By Symmetry Feb 12 at 14:06
1. Yes the raising operator $$b(\omega)^\dagger$$ corresponds to a mode of frequency omega and $$n(\omega) = b(\omega)^\dagger b(\omega)$$ count the number of excitations. You can see that the mode has a frequency of $$\omega$$ because that is the prefactor of the mode in the Hamiltonian, so one excitation of that mode increases the energy by $$\omega$$.