- #1
flyusx
- 65
- 10
- Homework Statement
- Two identical spin-half particles are confined to a one-dimensional box potential of size with walls at ##x=0## and ##x=L##.
(a) Find the ground state energy and the first excited state energy and their respective wave functions for this system when the two particles do not interact.
(b) Consider now that there exists a weakly attractive potential between the two particles. $$V_{p}=\begin{cases}-V_{0}&0\leq x\leq\frac{L}{2}\\0&\frac{L}{2}<x\leq L\end{cases}$$ Find the ground state and first excited state energies to first-order perturbation theory.
- Relevant Equations
- Nondegenerate Perturbation:
Degenerate Perturbation: Diagonalise in a degenerate subspace. The first-order energy corrections are the eigenvalues.
I am a bit confused in Part (b) in calculating the first-order perturbative first excited state energies.
I know that the total wavefunction must be antisymmetric because it is describing a system of fermions. This requires that an antisymmetric spatial wavefunction ##\phi## be paired with a symmetric spin state ##\chi## or an antisymmetric spin state be paired with a symmetric spatial state. In coupling two spin-half angular momenta, the spin triplet ##\left\vert s,m_s\right\rangle=\left\vert 1,\pm1\right\rangle,\left\vert 1,0\right\rangle## are symmetric while the spin singlet ##\left\vert0,0\right\rangle## is antisymmetric. A symmetric spatial wavefunction takes the form $$\phi_{s}=\frac{1}{\sqrt{2}}\left(\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)+\phi_{n_{1}}\left(x_{2}\right)\phi_{n_{2}}\left(x_{1}\right)\right)$$ and an antisymmetric spatial wavefunction takes the form $$\phi_{a}=\frac{1}{\sqrt{2}}\left(\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)-\phi_{n_{1}}\left(x_{2}\right)\phi_{n_{2}}\left(x_{1}\right)\right)$$ where ##\phi_{n}(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)## are the infinite potential well eigenstates.
The ground state has ##n_{1}=n_{2}=1## for a total energy ##E_{\text{GS}}=\frac{\pi^{2}\hbar^{2}}{mL^{2}}##. Because ##n_{1}=n_{2}##, the antisymmetric spatial wavefunction vanishes. The ground state wavefunction must therefore have a symmetric spatial part and an antisymmetric spin part. $$\psi_{\text{GS}}=\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)\left\vert0,0\right\rangle=\frac{2}{L}\sin\left(\frac{\pi x_{1}}{L}\right)\sin\left(\frac{\pi x_{2}}{L}\right)\left\vert0,0\right\rangle$$This state is nondegenerate, hence its first-order energy correction using perturbation theory is just the expectation value of ##\hat{V}_{p}## with respect to ##\phi_{\text{GS}}##.
The first excited state has one particle in ##n=1## and the other in ##n=2## for a total energy of ##E_{\text{FES}}=\frac{5\pi^{2}\hbar^{2}}{2mL^{2}}##. I came up with four wavefunctions corresponding to the first excited state.
$$\psi_{\text{FES},1}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,1\right\rangle$$$$\psi_{\text{FES},2}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,0\right\rangle$$$$\psi_{\text{FES},3}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,-1\right\rangle$$$$\psi_{\text{FES},4}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)+\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert0,0\right\rangle$$
That is, pairing each of the symmetric spin singlets with the antisymmetric spatial wavefunction and the antisymmetric spin singlet with the symmetric spatial wavefunction.
Because the first excited state is quadruply-degenerate, I started towards degenerate perturbation theory. In particular, I know I have to calculate (in theory) sixteen matrix elements corresponding to ##\left\langle\psi_{\text{FES},j}\vert\hat{V}_{p}\vert\psi_{\text{FES},k}\right\rangle## for ##j,k=1,2,3,4##. My point of confusion is that because the perturbative potential is only in position space, it leaves the spin states unaffected. Therefore, I got that all the off-diagonal matrix elements vanish from spin state orthonormality. $$\left\langle s',m_{s}'\vert s,m_{s}\right\rangle=\delta_{s',s}\delta_{m_{s}',m_{s}}$$ In other words, I would expect my perturbative Hamiltonian to be diagonal. In this case, what would the point of using degenerate perturbation theory be if the resulting matrix to be diagonalised is already diagonal? Should I expect this as a general case, ie whenever I am working with particles with spin and my perturbative Hamiltonian only acts on spatial wavefunction components, my perturbative Hamiltonian matrix is diagonal?
I know that the total wavefunction must be antisymmetric because it is describing a system of fermions. This requires that an antisymmetric spatial wavefunction ##\phi## be paired with a symmetric spin state ##\chi## or an antisymmetric spin state be paired with a symmetric spatial state. In coupling two spin-half angular momenta, the spin triplet ##\left\vert s,m_s\right\rangle=\left\vert 1,\pm1\right\rangle,\left\vert 1,0\right\rangle## are symmetric while the spin singlet ##\left\vert0,0\right\rangle## is antisymmetric. A symmetric spatial wavefunction takes the form $$\phi_{s}=\frac{1}{\sqrt{2}}\left(\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)+\phi_{n_{1}}\left(x_{2}\right)\phi_{n_{2}}\left(x_{1}\right)\right)$$ and an antisymmetric spatial wavefunction takes the form $$\phi_{a}=\frac{1}{\sqrt{2}}\left(\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)-\phi_{n_{1}}\left(x_{2}\right)\phi_{n_{2}}\left(x_{1}\right)\right)$$ where ##\phi_{n}(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)## are the infinite potential well eigenstates.
The ground state has ##n_{1}=n_{2}=1## for a total energy ##E_{\text{GS}}=\frac{\pi^{2}\hbar^{2}}{mL^{2}}##. Because ##n_{1}=n_{2}##, the antisymmetric spatial wavefunction vanishes. The ground state wavefunction must therefore have a symmetric spatial part and an antisymmetric spin part. $$\psi_{\text{GS}}=\phi_{n_{1}}\left(x_{1}\right)\phi_{n_{2}}\left(x_{2}\right)\left\vert0,0\right\rangle=\frac{2}{L}\sin\left(\frac{\pi x_{1}}{L}\right)\sin\left(\frac{\pi x_{2}}{L}\right)\left\vert0,0\right\rangle$$This state is nondegenerate, hence its first-order energy correction using perturbation theory is just the expectation value of ##\hat{V}_{p}## with respect to ##\phi_{\text{GS}}##.
The first excited state has one particle in ##n=1## and the other in ##n=2## for a total energy of ##E_{\text{FES}}=\frac{5\pi^{2}\hbar^{2}}{2mL^{2}}##. I came up with four wavefunctions corresponding to the first excited state.
$$\psi_{\text{FES},1}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,1\right\rangle$$$$\psi_{\text{FES},2}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,0\right\rangle$$$$\psi_{\text{FES},3}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)-\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert1,-1\right\rangle$$$$\psi_{\text{FES},4}=\frac{1}{\sqrt{2}}\left(\phi_{1}\left(x_{1}\right)\phi_{2}\left(x_{2}\right)+\phi_{2}\left(x_{1}\right)\phi_{1}\left(x_{2}\right)\right)\left\vert0,0\right\rangle$$
That is, pairing each of the symmetric spin singlets with the antisymmetric spatial wavefunction and the antisymmetric spin singlet with the symmetric spatial wavefunction.
Because the first excited state is quadruply-degenerate, I started towards degenerate perturbation theory. In particular, I know I have to calculate (in theory) sixteen matrix elements corresponding to ##\left\langle\psi_{\text{FES},j}\vert\hat{V}_{p}\vert\psi_{\text{FES},k}\right\rangle## for ##j,k=1,2,3,4##. My point of confusion is that because the perturbative potential is only in position space, it leaves the spin states unaffected. Therefore, I got that all the off-diagonal matrix elements vanish from spin state orthonormality. $$\left\langle s',m_{s}'\vert s,m_{s}\right\rangle=\delta_{s',s}\delta_{m_{s}',m_{s}}$$ In other words, I would expect my perturbative Hamiltonian to be diagonal. In this case, what would the point of using degenerate perturbation theory be if the resulting matrix to be diagonalised is already diagonal? Should I expect this as a general case, ie whenever I am working with particles with spin and my perturbative Hamiltonian only acts on spatial wavefunction components, my perturbative Hamiltonian matrix is diagonal?