Energy difference between symmetric and antisymmetric wavefunctions Is there any energy difference between a particle in a symmetric wavefunction and an identical particle in an identical potential but in a state with an anti-symmetric wavefunction?  Or is it case-dependent?
The full problem statement (from this assignment) is

 A: If we stick to the Problem #2 you mentioned, then yes, there is an energy difference. Since seems there is no degeneracy from symmetry for $ \psi(r-a) \pm \psi(r+a)$, unless numerically accident. 
$$ E_{\pm} = \langle \psi(r-a) \pm \psi(r+a)  | \hat{H} | \psi(r-a) \pm \psi(r+a) \rangle \tag{1} $$
$+,-$ correspond to $E_G,E_E$ in the Problem #2, respectively.
$E_G-E_E = 2 \left[ \langle \psi(r-a) | \hat{H} | \psi(r+a) \rangle + c.c. \right] $
$ =   2 \left[ \langle \psi(r-a) | \hat{H}_0 + \hat{H}'| \psi(r+a) \rangle + c.c. \right] $
$ =   2 \left[ \langle \psi(r-a) | E_0 + \hat{H}'| \psi(r+a) \rangle + c.c. \right] $
$ = 2 \left[ \langle \psi(r-a) |  \hat{H}'| \psi(r+a) \rangle + c.c. \right]  $
$ \tag{2} $
Here 
$$H_0:=-\frac{1}{2} \nabla^2 - \frac{1}{|r-a|}, H'= - \frac{1}{|r+a|} + \frac{1}{|2a|} $$ 
$c.c.$ denotes to complex conjugation. $E_0$ is the energy of hydrogen atom. 
we have assumed $\langle \psi(r-a) | \psi(r+a) \rangle=0$ as mentioned in the Problem #2. Since in the last line of Eq. (2), $\langle \psi(r-a) | 1/2a |\psi(r+a) \rangle=1/2a \langle \psi(r-a) | \psi(r+a) \rangle = 0$, we have $E_G-E_E<0$ 
