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Eigenstate of energy but not angular momentum?

  1. Nov 27, 2013 #1
    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:

    \left | \Psi \right > = c_1\left |n,l_1,m_1 \right > + c_2\left |n,l_2,m_2 \right>

    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.
  2. jcsd
  3. Nov 27, 2013 #2


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    Staff: Mentor

    You can choose a basis that has no eigenstates of any relevant operator - but there is no reason to do so, it just makes things more complicated. Eigenstates are nice, so bases are usually chosen to include the eigenstates for as many relevant operators as possible.
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