You have six shelves, one above the other and all above the floor, and six volumes of an encyclopedia, A, B, C, D, E and F.

(a) list all the ways you can arrange the volumes with five on the floor and one on the sixth/top shelf. One way might be$(ABCDE\_,\_,\_,\_,\_F)$.

(b) List all the ways you can arrange them with four on the floor and two on the third shelf.

(c) Show that there are many more ways, relative to pans (a) and (b), to arrange the six volumes with two on the floor and two each on the first and second shelves. (There are several ways to answer

this, but even listing them all won't take forever it's fewer than.)

(d) Suddenly, a fantastic change! All six volumes are volume X-it's impossible to tell them apart. For each of the three distributions described in parts (a), (b), and (c), how many different (distinguishable) ways are there now?

(e) If the energy you expend to lift a volume from the floor is proportional to a shelf's height, how do the total energies of distributions (a), (b), and (c) compare?

(I) Use these ideas to argue that the relative probabilities of occupying the lowest energy states should be higher for hosons than for classically distinguishable particles.

(g) Combine these ideas with a famous principle to argue that the relative probabilities of occupying the lowest states should he lower for fermions than for classically distinguishable particles.

Binomial coefficient:

The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number.

(a)

In this part, our task is to enumerate all possible ways we could arrange the books with five on the floor and one on the sixth shelf. To do so, let's first calculate the total number of ways for such an arrangement using the binomial coefficient:

$\begin{array}{rcl}W& =& \frac{6!}{5!1!}\\ & =& 6\\ & & \end{array}$

There are a total of six possible ways to have five books on the floor and one on the sixth shelf. The possible arrangements are as follows:

$$\begin{gathered} (ABCDE\_,\_,\_,\_,\_F) \\ (ABCDF,\_,\_,\_,\_,\_,E) \\ (ABCEF,\_,\_,\_,\_,\_,D) \\ (ABDEF,\_,\_,\_,\_,\_,C) \\ (ACDEF,\_,\_,\_,\_,\_B) \\ (BCDEF,\_,\_,\_,\_,\_,A) \end{gathered}$$

(b)

In this part, our task is to list all possible ways we could arrange the books such that four are on the floor, and two are on third shelf. Again, let us use the binomial coefficient to list down the total ways to perform such an arrangement:

$\begin{array}{rcl}W& =& \frac{6!}{4!2!}\\ & =& 15\\ & & \end{array}$

There are a total of 15 possible ways to have four books on the floor and two on the third shelf. The possible arrangements are as follows:

$$\begin{gathered} \left(ABCD,\_,\_,EF,\_,\_,\_\right);\left(BCEF,\_,\_,AD,\_,\_,\_\right) \\ \left(ABEF,\_,\_,CD,\_,\_,\_\right);(BDEF,\_,\_,AC,\_,\_,\_) \\ (CDEF,\_,\_,AB,\_,\_,\_);(ACEF,\_,\_,BD,\_,\_,\_) \\ (ABCE\_,\_,DF,\_,\_,\_);(ADEF\_,\_,BC,\_,\_,\_) \\ (ABDE,\_,\_,,CF,\_,\_,\_);ABDF,\_,\_,CE,\_,\_,\_ \\ (ACDE,\_,\_,,BF,\_,\_,\_);\left(ACDF,\_,\_,BE,\_,\_,\_\right) \\ (BCDE,\_,\_,AF,\_,\_,\_);\left(ABCF,\_,\_,DE,\_,\_,\_\right) \\ (BCDF,\_,\_,AE,\_,\_,\_) \end{gathered}$$

(c)

Using the binomial coefficient, the total number of ways to arrange the six volumes such that two are on the floor, two are on the first shelf, and two are on the second shelf is thus:

$\begin{array}{rcl}W& =& \frac{6!}{2!2!2!}\\ & =& 90\\ & & \end{array}$

(d)

Now supposing that the volumes cannot be distinguished from one another, that is, all books are seemingly identical, then our answers in parts (a), (b), and (c) are all one. This means that if the books are indistinguishable, then the number of ways to perform the arrangements described in parts (a), (b), and (c) is only one.

(e)

Supposing that the energy we expend to lift the books are proportional to the shelf height, let us now say that the spacing between energy levels, that is, between shelves, is one. Thus, on the floor, we have 0 units of energy to disburse, on the first shelf, we have 1 energy unit, on the second, we have 2 units, and so on.

Now for part (a), five books are on the floor, and one is on the sixth shelf. Hence the total energy we disbursed must be:

$\begin{array}{rcl}{E}_a& =& 0\left(5\right)+6\left(1\right)\\ & =& 6\text{units}\\ & & \end{array}$

For part (b), four books are on the floor and two are on the third shelf. The total energy expended is

$\begin{array}{rcl}{E}_b& =& 0\left(4\right)+3\left(2\right)\\ & =& 6\text{units}\\ & & \end{array}$

For part (c),

$\begin{array}{rcl}{E}_c& =& 0\left(2\right)+1\left(2\right)+2\left(2\right)\\ & =& 6\text{units}\\ & & \end{array}$

As we can see, all arrangements described in parts (a). (b). and (c) share the same total energy of 6 units.

(f)

As we've seen, books that can be recognised from one another have a greater number of alternative configurations than books that appear to be similar. This is similar to classical particles, which have a considerably larger variety of ways to disperse their energies in lower energy states than bosons. As a result, bosons gravitate toward lower energy levels, especially the ground state.

(g)

No two particles can have the same set of quantum numbers, according to the exclusion principle. As a result, if the spin of the particles is taken into account, the maximum number of fermions that can occupy a given energy state is two. Despite the fact that fermions are identical particles, the exclusion principle prevents particles from sharing energy levels. The number of ways they can divide their energies in the lowest levels alters as a result of this. As a result, fermions have a smaller likelihood of being found in the lowest states than classical particles