Problem about identical particles

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Lebnm
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Can someone help me with this exercise?: Consider two electrons constrained to move in one dimension. They interact through the potential
$$
V(x) = \begin{cases}
0, & \text{ if } |x| > a \\
-V_{0}, & \text{ if } |x| \leq a
\end{cases}
$$
where ##x## is the relative coordinate. The total spin is ##S = 1##, and the objective is to determine the state of the system.
I know I can write the state as ##|\Psi \rangle = |P \rangle \otimes |\psi \rangle \otimes |SM \rangle##, where ##|P \rangle## is an eigenstate of the momentum of CM, and the wave function ##\psi (x) = \langle x | \psi \rangle ## is the solution of $$-\frac{\hbar^{2}}{2\mu}\frac{\mathrm{d^{2}}\psi }{\mathrm{d} x^{2}} + V(x)\psi(x) = E \psi(x).$$ Now, I know that the states ##|P \rangle## and ##|SM \rangle ## with ##S=1## are symmetric over permutation of the electrons, but I don't know how to determine if ##|\psi \rangle## is either symmetric or antisymmetric. In this case, the permutation of the electrons change ##x \rightarrow -x##, so it has the same effect of parity. The hamiltnian above is obviously invariant over parity, and there is a theorem that says ##|\psi \rangle## is an eigenstate of parity (suposing that the spectrum of the hamiltonian is non-degenerated), so it have to be symmetic or antisymmetic. Do I need to determine the wave function ##\psi(x)## to it? I tried to do this, but I get four constants that I can't determine.
 
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Lebnm said:
Now, I know that the states ##|P \rangle## and ##|SM \rangle ## with ##S=1## are symmetric over permutation of the electrons, but I don't know how to determine if ##|\psi \rangle## is either symmetric or antisymmetric.
What does the Pauli principle tell you?
 
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Ok, I understood. The total state have to be antisymmetric, so ##| \psi \rangle ## is antisymmetric, since ##|P\rangle## and ##|S,M\rangle## are symmetric. Solving the problem for ##\psi(x)##, I find

$$
\psi(x) =
\begin{cases}
Ae^{ikx} + Be^{-ikx}, & \text{ if } |x| > a \\
Ce^{ilx} + De^{-ilx}, & \text{ if } |x| \leq a
\end{cases}
$$

where ##k = \sqrt{2\mu E}/\hbar##, ##l = \sqrt{2\mu(E+V_{0})}/\hbar## and ##E = E_{TOT} - E_{CM}##. Imposing that ##\psi(-x) = - \psi(x)##, we have ##A = -B## and ##C = - D##, so

$$
\psi(x) =
\begin{cases}
Asin(kx), & \text{ if } |x| > a \\
Csin(lx), & \text{ if } |x| \leq a
\end{cases}
$$

But ##\psi## also need to be continuous, what implies that ##C = A sin(ka)/sin(la)##, and ##A## have to be chosen such that ##\langle \Psi | \Psi \rangle = 1##. The orbital part of the wave function will be

$$
\Psi(x, X) \propto e^{iPX/\hbar}
\begin{cases}
sin(kx), & \text{ if } |x| > a \\
\frac{sin(ka)}{sin(la)}sin(lx), & \text{ if } |x| \leq a
\end{cases}
$$

where ##X## is the coordinate of C.M. Now I have to multiply this by the spin state. But I have three of them: ##|1,1 \rangle##, ##|1,0 \rangle## and ##|1,-1 \rangle##. Do I need to take a linear combination of them? In this case, my wave function would depend of three constants, is it correct?
 
Actually, this is a problem that I saw in the internet and I don't remember the exact text. But the potential is that ##V## above and the total spin is ##S = 1##. Is this enough to find the soluction?
 
Lebnm said:
Actually, this is a problem that I saw in the internet and I don't remember the exact text. But the potential is that ##V## above and the total spin is ##S = 1##. Is this enough to find the soluction?
No, far from it. At best, this is an exercise to understand the constraints on the spatial wave function.