- #1
Jerrynap
- 8
- 0
In a simple case of hydrogen, we can have simultaneous eigenstate of energy, angular momentum [itex] L_z, \hat{\vec{L}^2} [/itex]. I'm thinking of constructing a state that is an eigenstate of energy but not the angular momentum:
[itex]
\left | \Psi \right > = c_1\left |n,l_1,m_1 \right > + c_2\left |n,l_2,m_2 \right>
[/itex]
In this particular state, when I measure the energy, the state is left unchanged. So it is an energy eigenstate. However, when I measure the angular momentum, the state collapses into either [itex] l_1 [/itex] or [itex] l_2 [/itex], and subsequent measurements of the angular momentum and energy will leave the state unchanged. So the original state is not an eigenstate of angular momentum.
My question is that, we are always talking about simultaneous eigenstate of two commuting operators, but are we free to choose the set of basis that are eigenstates of one operator and not the other? This seems to be true only if we have degeneracies.
[itex]
\left | \Psi \right > = c_1\left |n,l_1,m_1 \right > + c_2\left |n,l_2,m_2 \right>
[/itex]
In this particular state, when I measure the energy, the state is left unchanged. So it is an energy eigenstate. However, when I measure the angular momentum, the state collapses into either [itex] l_1 [/itex] or [itex] l_2 [/itex], and subsequent measurements of the angular momentum and energy will leave the state unchanged. So the original state is not an eigenstate of angular momentum.
My question is that, we are always talking about simultaneous eigenstate of two commuting operators, but are we free to choose the set of basis that are eigenstates of one operator and not the other? This seems to be true only if we have degeneracies.