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i am studying for a physics exam and came across an exercise i cannot seem to crack. I have read through Feynmans lecture books & a lot of the internet but i am somewhat stuck. first the question, and then what i have got so far:

(Note the problem is originally in German, I hope my translation does not change the problem)

Problem Description: We look at 2 identical (indistinguishable) particles in a one-dimensional box with infinite potential energy at the bounds. The box' width is 2a. in the box the potential energy is 0 and on the bounds it is infinite. we take as granted that the particles DON'T interact.

1) Find the 4 lowest Energies the system can have.

2) How many states of the same energy does every of those 4 levels have if both particles have spin 1/2 (fermions)?

3) How many states of the same energy does every of those 4 levels have if both particles have spin 1 (bosons)?

Ok, now lets see how far I've come

1) We know, that the Energy levels of 1 particle in such a box of width 2a is given by

$$E_n = \frac{n^2h^2}{32ma^2}$, \quad\quad n=(n_1+n_2)$$

This gives me the following 4 states:

$$E_1 = \frac{h^2}{32ma^2},\quad E_2 = \frac{h^2}{8ma^2},\quad E_3= \frac{9h^2}{32ma^2},\quad E_4 = \frac{h^2}{2ma^2} $$

2) Here it gets tricky for me.Noting $\Phi(x)$ as the complete statefunction, $\Psi(x)$ for the Place and $\Xi(x)$ for the spin-state function and the subscripts $_s$ for symmetric and $_a$ for antisymmetric, we know that the complete function must be antisymmetrically for fermions. this is achieved by combining a symmetric place & antisymmetric spin or the other way around.

Only looking at $\Psi(x)$ for now, we find the following n-states for our 2 body system. noting $n := (n_1,n_2)


(1,0) with $\Psi_a$ and $\Psi_s$ (2 degeneracies)


(1,1), $\Psi_s$ (1 degeneracie)

(2,0) $\Psi_a$ and $\Psi_s$ (2 degeneracies)


(3,0) $\Psi_a$ and $\Psi_s$ (2 degeneracies)

(2,1) $\Psi_a$ and $\Psi_s$ (2 degeneracies)


(4,0) $\Psi_a$ and $\Psi_s$ (2 degeneracies)

(3,1) $\Psi_a$ and $\Psi_s$ (2 degeneracies)

(2,2) $\Psi_s$ (1 degeneracie)

Now i have a look at the spin-states $\Xi(x)$ for Spin = S = 1, using the formulas

Number of antisymm spinstates # $\Xi_a(x) = S(2S+1) = 1$

Number of symmetric spinstates # $\Xi_s(x) = (S+1)(2S+1) = 3$

Now we Combine those to antisymmetric functions using

$\Phi_a = \Psi_a+\Xi_s$ or $\Phi_a=\Psi_s+\Xi_a$

To count the states of the same energy i use this formula

N states = $( N\Psi_a \cdot N\Xi_s)+( N\Psi_s \cdot N\Xi_a)$ (N denoting the number of states using the degeneracy of each n-combination)

This finally results in the following degeneracies

$N E_1= 2*1 + 2*3 = 8$

$N E_2= 2*1 + 3*3 = 11$

$N E_3= 4*1 + 4*3 = 16$

$N E_4= 4*1 + 5*3 = 19$

3) Now the same for Bosons (Spin S = 1). In this case the complete function $\Phi$ must be symmetrically, so we have to combine antisym spin and antisym place or both symm.

The degeneracy of the place function stays the same as in 2), but for the spin we have to recalculate using:

Number of antisymm spinstates # $\Xi_a(x) = S(2S+1) = 3$


To achieve symmetric complete functions, we also need to adjust our N E formula which will be changed to :

N states = $( N\Psi_a \cdot N\Xi_a)+( N\Psi_s \cdot N\Xi_s)$ (N denoting the number of states using the degeneracy of each n-combination)

This finally results in

$N E_1= 2*6 + 2*3 = 18$

$N E_2= 2*6 + 3*3 = 21$

$N E_3= 4*6 + 4*3 = 36$

$N E_4= 4*6 + 5*3 = 39$

Conclusion and MY question: Sadly this exercise does not have a solution, so I decided to post my attempt here and hope that some of you might find any errors I made and help me if they spot if and where my thoughts went rogue.

Hoping for some good input! (Please add questions if something is not clear - it could be the translation).

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thx for editing, altough: it is definitly NOT homework. its an exercise from an old exam. (at a certain point there is no such thing as 'homework' anymore) ;) –  Sebastian Flückiger Jul 20 '12 at 8:38
Hi Sebastian - generally we discourage questions that just ask for someone to check your work. This site is meant to be for conceptual questions, so if there's some specific concept confusing you, it's fine to ask about that. –  David Z Jul 20 '12 at 23:52
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closed as too localized by David Z Jul 20 '12 at 23:49


2 Answers

up vote 2 down vote accepted

The way you would manually find the wave functions is more complicated than this. At first, you would need to describe better the one-particle scenario. I don't really understand your concept of the spin-state function, so I'll use mine, I hope it won't make my answer unusable.

One particle

If the particle has no spin, the wave functions are completely determined by energy. In other words, the energy levels $E_n$ are nondegenerate. We can assign wave functions, say $\psi_n(x)$, to them directly.

If we add spin to the description of the particle, it does not change much in our observation of energy, but the underlying description changes. Namely, we can not get on with simple $\mathbb{R} \to \mathbb{C}$ functions. There are actually two equivalent concepts:

  1. defining a spinor wave function $\Psi: \mathbb{R} \to \mathbb{C}^d$, mapping each point in space to a $d$-tuple of probability amplitudes, where $d$ is the dimension of the spin space, $d = 2s+1$, and
  2. making the spin state another independent variable beside position. In this way, the wave function would be defined as $\psi(x,\sigma): \mathbb{R} \times S \to \mathbb{C}$. Here, $\sigma$ takes the values from $\{-s, -s+1, \ldots, s-1, s\}$ (which makes $2s-1$ values) and gives a single probability amplitude, that of finding the particle at position $x$ and with instantaneous spin state $\sigma$. The equivalence is given by the fact that these probability amplitudes are simply the components of the spinor $\Psi(x)$: $$\Psi(x) = \begin{pmatrix}\psi(x,-s) \\ \vdots \\ \psi(x,s-1) \\ \psi(x,s)\end{pmatrix}$$

So, if we assume a particle with spin, say, $\frac12$, all the energy levels become twice degenerated, because for energy $E_n$, we can find two linearly independent wave functions: $$\Psi_{n, \uparrow}(x) = \begin{pmatrix}\psi_n(x) \\ 0\end{pmatrix} \quad \hbox{and} \quad \Psi_{n, \downarrow}(x) = \begin{pmatrix}0 \\ \psi_n(x)\end{pmatrix},$$ reusing the scalar $\psi_n$ single-particle wavefunctions. These are joint eigenfunctions of the Hamiltonian and spin operators: "energy $E_n$ spin up" and "energy $E_n$ spin down".

Two particles

When we consider two noninteracting particles, the solution further depends on whether they are distinguishable or not. Let's briefly address distinguishable particles, despite the fact that you are not asking about them, but for future reference.

In either case, the state space expands to cover all combinations of both the particles' positions and spin states. Therefore, in the formalism I called "2.", we get wave functions indexed by four coordinates: continuous variables $x$, $y$ and discrete variables $\sigma$, $\tau$.

If the particles can be distinguished, we can build those functions as tensor products of one-particle functions, i.e.: $$\psi^{(2)}(x,y,\sigma,\tau) = \psi_a(x,\sigma) \psi_b(y,\tau),$$ and as superpositions thereof: $$\psi^{(2)}(x,y,\sigma,\tau) = \sum_{i=1}^k \alpha_i \psi_{a_i}(x,\sigma) \psi_{b_i}(y,\tau).$$

In the case of indistinguishable particles, we must pay extra attention in order for the result to be symmetric or antisymmetric wrt. exchange of $(x,\sigma) \leftrightarrow (y,\tau)$. We usually do so by constructing the functions using permutations in the first place, for example, $$\psi^{(2,symm.)}(x,y,\sigma,\tau) = \psi_a(x,\sigma) \psi_b(y,\tau) + \psi_a(y,\tau) \psi_b(x,\sigma),$$ $$\psi^{(2,asymm.)}(x,y,\sigma,\tau) = \psi_a(x,\sigma) \psi_b(y,\tau) - \psi_a(y,\tau) \psi_b(x,\sigma).$$ In more than 2 particles, this takes a lot of work (and storage) because one has to go through all the permutations, but is doable.

Now, for each of your sub-questions, you would need to find the combinations of state assignments to "a" and "b" which give the energies you need, keeping on mind that permutations mean no difference in the state, and that $a$ must not equal $b$ if the exclusion principle is in action (fermions). The count of these would be the degeneracies. Luckily, given either of the 4 lowest energies of the form $E_{n_1} + E_{n_2}$, the decomposition to the sum of two single-particle energies is unique, so we only need to care about spin.

For example, the energy $2E_2$ can be reached for two $\frac12$ particles only one way, $a$ = energy $E_2$, spin up, $b$ = energy $E_2$, spin down (or vice versa, which gives the same quantum state). Thus this energy level is nondegenerate. On the contrary, energy $E_1+E_2$ is an eigenvalue for a total of four states:

  1. $a$ = energy $E_1$, spin up, $b$ = energy $E_2$, spin up,
  3. $a$ = energy $E_1$, spin down, $b$ = energy $E_2$, spin up.
  4. $a$ = energy $E_1$, spin down, $b$ = energy $E_2$, spin down.

In the case of bosons, the Pauli exclusion principle goes away, giving us more possibilities. For example, energy $2E_2$ is obtained if

  1. $a$ = energy $E_2$, spin "+1", $b$ = energy $E_2$, spin "+1",
  2. $a$ = energy $E_2$, spin "+1", $b$ = energy $E_2$, spin "0",
  3. $a$ = energy $E_2$, spin "+1", $b$ = energy $E_2$, spin "-1",
  4. $a$ = energy $E_2$, spin "0", $b$ = energy $E_2$, spin "0",
  6. $a$ = energy $E_2$, spin "-1", $b$ = energy $E_2$, spin "-1",

The degeneracy of this level is 6.

Another way

As the above can get daunting after a while, we can use a simpler method if only the energy levels and degeneracies are the question, not the wave functions themselves. This method uses the occupation diagrams well-known from chemistry (which are mere graphical representations of multi-particle wave functions).

In the case of spin $\frac12$, we prepare boxes corresponding to single-particle energies $E_1$, $E_2$, $E_3$, and $E_4$. We can put in two fermions in the following ways:

  • $(\uparrow \downarrow), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$, degeneracy 1
  • $(\uparrow \ ), (\uparrow\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(\uparrow \ ), (\downarrow\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(\downarrow \ ), (\uparrow\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(\downarrow \ ), (\downarrow \ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$ ... degeneracy 4,
  • $(\ \ ), (\uparrow \downarrow), (\ \ ), (\ \ )$ ... energy $2E_1$, degeneracy 1,

etc. Now you only have to find which energies are the lowest four, anyway, it becomes quickly obvious that the degeneracy only depends on the fact whether they are in the form of $2E_n$ for some $n$ or $E_m+E_n$ for some $m \ne n$.

For spin $1$ bosons, the spin state can become "+1", "0", or "-1" and the particles don't obey the exclusion principle, so we don't need to care about putting two like "arrows" in the same box. Similarly as above, the order within one box does not make a difference. Let us save same space by using abbreviations "+" / "0" / "-" for the spin states. Here we go:

  • $(++), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$,
  • $(+0), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$,
  • $(+-), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$,
  • $(00), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$,
  • $(0-), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$,
  • $(--), (\ \ ), (\ \ ), (\ \ )$ ... energy $2E_0$ ... degeneracy 6,
  • $(+\ ), (+\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(+\ ), (0\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(+\ ), (-\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(0\ ), (+\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(0\ ), (0\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(0\ ), (-\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(-\ ), (+\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(-\ ), (0\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$,
  • $(-\ ), (-\ ), (\ \ ), (\ \ )$ ... energy $E_0+E_1$ ... degeneracy 9,

and so on. Every energy level consisting of a sum of two different energies get a degeneracy of 9, or 6 if the two components are the same.

Final remark: Under further inspection, it can be found that the "general observations" break at higher energies, where the Diophantine equation $m^2 + n^2 = a$ starts having multiple solutions, e.g. the energy $260\frac{h^2}{32ma^2}$ could be $\frac{h^2}{32ma^2}(16^2+2^2)$ as well as $\frac{h^2}{32ma^2}(14^2+8^2)$. You don't need to worry about this in the beginning of the spectrum, though.

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thank you for this unbelievable thourough answer! it will take some time for me to process and understand it i will come back on it :) –  Sebastian Flückiger Jul 20 '12 at 12:02
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Your answer to 1 is correct. The rest is wrong. For two spin 1/2 fermions, there is only one states at each level, the two fermions must be in an antisymmetric spin state, regardless of the level. This means that they are forced to have opposite spin to be at this level. For spin 1, there are 6 symmetric spin states per level, the two spin states make a spin 2 symmetric and a spin 0 symmetric spin wavefunction.

The reason you are getting confused is because you are confusing the parity symmetry of the state under reflections with the exchange symmetry under interchange of identical particles. The two symmetries are completely separate. The particles don't get spatially reflected when you interchange them, the symmetry or antisymmetry of the state in the spatial sense is irrelevant.

If the two fermions felt a potential toward each other, as opposed to moving in an external potential, then your analysis would be correct. In this case, exchanging the two particles would correspond to reflecting the relative coordinate between them. That would make a less trivial exam problem. But that's not the problem you were given.

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