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:

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