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I What's an eigenstate

  1. May 29, 2017 #1
    What is an eigenstate in relation to the Schodinger equation?

    We've been working with this stuff but I don't exactly understand what that is.
    I know of linear algebra eigenstates or eigenfunctions but I don't know if they are directly related.
     
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  3. May 29, 2017 #2

    andrewkirk

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    They are directly related. The set of all eigenstates of a non-degenerate linear operator form a basis for the vector space, which in the Schrodinger case is the vector space of all possible configurations of the system.

    In the Schrodinger equation, we can use this basis property to express the initial state as a linear sum of eigenstates of a particular linear operator. The Schrodinger equation then tells us that each of those evolves over time by the scalar multiplier ##e^{ikt}## changing with time ##t##. The scalar multiplier is the 'phase'. Because ##k## differs between different eigenstates, the periods for phase variation vary between eigenstates, so the relative phase changes, which means that the overall state, which is the sum of the individual components, varies over time in a way that is not just a scalar multiplier.
     
  4. May 29, 2017 #3

    jtbell

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    If you're working with the Schrödinger equation, you're probably working with wavefunctions ##\psi(x,t)## or ##\psi(x)##.

    If ##\psi## is an eigenstate of an operator, then when you apply the operator to ##\psi##, you get ##\psi## multiplied by a constant (the eigenvalue).

    For example, if ##\psi## is an eigenstate of the momentum operator ##-i\hbar \frac {\partial} {\partial x}##, then $$-i\hbar \frac {\partial \psi} {\partial x} = p\psi$$ where ##p## is the eigenvalue, i.e. the momentum of that state.
     
  5. May 30, 2017 #4
  6. May 30, 2017 #5
    So more or less, eigenstates are simply solutions?
     
  7. May 30, 2017 #6
    Yes. They are solutions to the eigenvalue-problem of the operator corresponding to the observable in question.
     
  8. May 30, 2017 #7

    PeroK

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    Not really. Energy Eigenstates are solutions to the time independent Schrödinger equation, each for a corresponding energy value.

    But, more generally, any linear operator has its eigenstates. It is an assumption in QM that any operator that represents an observable has a complete set of orthogonal eigenstates. The Hamiltonian being a specific example of this.

    Other operators may have eigenstates with different properties. The raising and lowering operators of the harmonic oscillator are good examples. The eigenstates of the lowering operator are the so-called coherent states. And the raising operator has no eigenstates.

    PS eigenstates relate directly to linear operators are are really only indirectly related the Schrödinger equation itself.
     
    Last edited: May 30, 2017
  9. May 30, 2017 #8
    Ok but it seems like you just said eigenstates are solutions. YOu said that any linear operator has its eigenstates so those eigenstates are simply "solutions" corresponding to specific quantum numbers right?
     
  10. May 30, 2017 #9

    PeroK

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    Nothing is a "solution". Things are solutions to specific equations.

    Energy eigenstates are solutions to the time independent Schrödinger equation.

    Energy eigenstates are not solutions to the full Schrödinger equation. For that you need to include a function of time, related to the energy.

    An eigenstate of a linear operator is well defined without any reference whatsoever to the Schrödinger equation or to QM for that matter. It is defined as a solution to the eigenvalue equation for that operator.
     
  11. May 30, 2017 #10

    jtbell

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    To make this more explicit, the operator in question here is the Hamiltonian operator: $$\hat H = -\frac {\hbar^2} {2m} \frac {\partial^2} { \partial x^2} + V(x)$$ Written in the form of an eigenvalue equation, the Schrödinger equation is ##\hat H \psi = E \psi##.
     
  12. May 30, 2017 #11

    dextercioby

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    This is a self-contradiction. Can you, please, re-phrase your opinion?
     
  13. May 30, 2017 #12

    andrewkirk

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    I think the sentence just inadvertently lacks the word 'Hermitian'. Any Hermitian operator has a complete basis of eigenstates. The raising operator is not Hermitian.
     
  14. May 31, 2017 #13

    PeroK

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    Any linear operator has a set of eigenstates. That set may be empty.
     
  15. May 31, 2017 #14

    dextercioby

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    Yes, the raising operator is not Hermitean, but this doesn't mean it has no eigenstates.
     
  16. May 31, 2017 #15

    andrewkirk

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    Sure, I didn't say it did. But with the insertion of 'Hermitian' in the post, it is not self-contradictory, which is what was being discussed.
     
  17. May 31, 2017 #16

    dextercioby

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    Alright, putting "hermitean" there is a step forward as the first statement is rephrased. The raising operator has eigenstates, it's then the second statement of the two I quoted that is wrong.
     
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