6.4 Ground State of a System of Bosons

The ground state for a system of non­interacting spinless bosons is simple. The ground state is defined as the state of lowest energy, so every boson has to be in the single-particle state $\pp111/{\skew0\vec r}///$ of lowest energy. That makes the system energy eigen­function for spinless bosons equal to:

$$\psi_{\rm gs,\ bosons} = \pp111/{\skew0\vec r}_1/// \tim... ...}_2/// \times \ldots \times \pp111/{\skew0\vec r}_I/// %$$ (6.8)

If the bosons have spin, this is addi­tionally multi­plied by an arbitrary combin­ation of spin states. That does not change the system energy. The system energy either way is $I{\vphantom' E}^{\rm p}_{111}$, the number of bosons times the single-particle ground state energy.

Figure 6.2: Ground state of a system of noninter­acting bosons in a box.

Graphi­cally, the single-particle ground state $\pp111////$ is the point closest to the origin in wave number space. It is shown as a fat blue dot in figure 6.2 to indicate that all $I$ bosons are bunched together in that state.

Physicists like to talk about “occupation numbers.” The occupation number of a single-particle state is simply the number of particles in that state. In particular, for the ground state of the system of non­interacting spinless bosons above, the single-particle state $\pp111////$ has occupation number $I$, while all other single-particle states have zero.

Note that for a macroscopic system, $I$ will be a humongous number. Even a millimol of particles means well over 10$\POW9,{20}$ particles. Bosons in their ground state are very unfair to the single-particle states: $\pp111////$ gets all of them, the rest gets nothing.

Key Points

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For a system of bosons in the ground state, every boson is in the single particle state of lowest energy.