- #1
logic smogic
- 54
- 0
[SOLVED] Perturbed Ground State Wavefunction with Parity
A particle is in a Coulomb potential
H_{0}=\frac{|p|^{2}}{2m} - \frac{e^{2}}{|r|}
When a perturbation V (which does not involve spin) is added, the ground state of H_{0} + V may be written
|\Psi_{0}\rangle = |n=0,l=0,m=0\rangle + \sum C_{nlm}|n,l,m \rangle
where |n,l,m\rangle is a Hyrdogenic wavefunction with radial quantum number n and angular momentum quantum numbers l, m.
Consider the following possible symmetries of the perturbation V. What constraints, if any, would the presence of such a symmetry place on the possible values of the coefficients C_{nlm}?
(i) Symmetry under Parity.
(ii) Rotational symmetry about the z axis.
(iii) Full rotational symmetry.
(iv) Time reversal symmetry.
Let's stick to just part (i) for now.
Initially, I want to understand what the question is asking, and what that formula for the ground state ket means. Here's what went through my head:
1. Time-Independent perturbation theory tells us that the eigenkets can be written:
|n \rangle = |n_{0} \rangle + \sum |m_{0} \rangle \langle m_{0} | n \rangle
where,
\langle m_{0} | n \rangle = \lambda \frac{\langle m_{0} | V | n \rangle}{E_{n} - E_{m,0}}
2. Comparing this to the equation of the perturbed ground state (in the statement of the problem), it looks like,
|n=0,l=0,m=0\rangle \rightarrow |n_{0} \rangle
and,
|n,l,m \rangle \rightarrow |m_{0} \rangle
which leaves the "coefficients" as,
C_{nlm} \rightarrow \langle m_{0} | n \rangle
3. As for parity, I know that if \pi is the parity operator, then for an operator A,
\pi^{\dagger} A \pi = \pm A
Presumably, I should apply this to V, and learn something about the possible values of the coefficients.
This is where I'm stuck. I've no idea how this is supposed to restrict the values that C_{nlm} can take. Any suggestions or hints in the right direction?
Homework Statement
A particle is in a Coulomb potential
H_{0}=\frac{|p|^{2}}{2m} - \frac{e^{2}}{|r|}
When a perturbation V (which does not involve spin) is added, the ground state of H_{0} + V may be written
|\Psi_{0}\rangle = |n=0,l=0,m=0\rangle + \sum C_{nlm}|n,l,m \rangle
where |n,l,m\rangle is a Hyrdogenic wavefunction with radial quantum number n and angular momentum quantum numbers l, m.
Consider the following possible symmetries of the perturbation V. What constraints, if any, would the presence of such a symmetry place on the possible values of the coefficients C_{nlm}?
(i) Symmetry under Parity.
(ii) Rotational symmetry about the z axis.
(iii) Full rotational symmetry.
(iv) Time reversal symmetry.
The Attempt at a Solution
Let's stick to just part (i) for now.
Initially, I want to understand what the question is asking, and what that formula for the ground state ket means. Here's what went through my head:
1. Time-Independent perturbation theory tells us that the eigenkets can be written:
|n \rangle = |n_{0} \rangle + \sum |m_{0} \rangle \langle m_{0} | n \rangle
where,
\langle m_{0} | n \rangle = \lambda \frac{\langle m_{0} | V | n \rangle}{E_{n} - E_{m,0}}
2. Comparing this to the equation of the perturbed ground state (in the statement of the problem), it looks like,
|n=0,l=0,m=0\rangle \rightarrow |n_{0} \rangle
and,
|n,l,m \rangle \rightarrow |m_{0} \rangle
which leaves the "coefficients" as,
C_{nlm} \rightarrow \langle m_{0} | n \rangle
3. As for parity, I know that if \pi is the parity operator, then for an operator A,
\pi^{\dagger} A \pi = \pm A
Presumably, I should apply this to V, and learn something about the possible values of the coefficients.
This is where I'm stuck. I've no idea how this is supposed to restrict the values that C_{nlm} can take. Any suggestions or hints in the right direction?
Last edited: