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Permutation for classical particle $$\Omega=\frac{N!}{\Pi n_i!}$$ By using Lagrange method of undetermined multiplier, we get $$n_i=Ae^{\frac{-E}{kT}}$$ Probability, $$p=\frac{n_i} {Z}$$ where we define $Z$ as $$Z_{classical}≡∑e^{\frac{-E}{kT}}$$

But for fermion, $$n_i=(Ae^{\frac{E}{kT}}+1)^{-1}$$ My question is: why do we have the same partition formula for classical and quantum particle? If possible, please tell me how to define the partition function for fermion. Thanks

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That definition of $\Omega$ only works if you have distinguishable particles and you can put as many as you want in any state. That gives you the Maxwell-Boltzmann distribution.

In quantum mechanics, particles of the same type are not distinguishable; e.g. there's no way to tell one electron from another. Moreover, for Fermions (like electrons), you can have at most one particle in a given state. Bosons (like photons) can have as many particles in one state as you please.

These give rise to two '$\Omega$'s, both different than the one you gave.

As for the partition function, it takes the same form, but you sum over different things. Define

$$Z=\sum_se^{-E_s\beta}$$

where the sum is over the possible microstates of the system. (A state is an energy level that a particle can occupy; e.g. the ground state of a hydrogen atome. A microstate is a configuration of all the particles in the system; some examples are given below.)

Say you have three particles and three states available to you. For Bosons, you have many different microstates available to you: all three particles in state one, one particle in each state, all three particles in state three, etc. So, your sum has many terms. For Fermions, you have only one option: one particle in each state, since there are three particles, three states, and only one particle can be in each state. So there is only one microstate available for Fermions in this system, and thus the sum only has one term.

Short version: the equation for the partition function stays the same, but the microstates you can sum over are very different.

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