What is an Eigenstate in Relation to Schrodinger Equation?

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In summary, an eigenstate in relation to the Schrödinger equation is a solution to the eigenvalue equation for a specific linear operator. In the Schrödinger equation, the initial state can be expressed as a linear combination of eigenstates, each of which evolves over time with a scalar multiplier. These eigenstates are not only applicable to the Hamiltonian operator, but to any Hermitian operator. However, the raising operator is an example of a non-Hermitian operator that still has eigenstates.
  • #1
CookieSalesman
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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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  • #2
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.
 
  • #3
CookieSalesman said:
What is an eigenstate in relation to the Schodinger equation?
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.
 
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thx guys
 
  • #5
So more or less, eigenstates are simply solutions?
 
  • #6
CookieSalesman said:
So more or less, eigenstates are simply solutions?
Yes. They are solutions to the eigenvalue-problem of the operator corresponding to the observable in question.
 
  • #7
CookieSalesman said:
So more or less, eigenstates are simply solutions?

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.
 
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  • #8
PeroK said:
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.
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?
 
  • #9
CookieSalesman said:
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?

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.
 
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  • #10
PeroK said:
Energy eigenstates are solutions to the time independent Schrödinger equation.
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##.
 
  • #11
PeroK said:
[...] But, more generally, any linear operator has its eigenstates. [...] And the raising operator has no eigenstates.[...]

This is a self-contradiction. Can you, please, re-phrase your opinion?
 
  • #12
dextercioby said:
This is a self-contradiction. Can you, please, re-phrase your opinion?
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.
 
  • #13
dextercioby said:
This is a self-contradiction. Can you, please, re-phrase your opinion?

Any linear operator has a set of eigenstates. That set may be empty.
 
  • #14
andrewkirk said:
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.

Yes, the raising operator is not Hermitean, but this doesn't mean it has no eigenstates.
 
  • #15
dextercioby said:
Yes, the raising operator is not Hermitean, but this doesn't mean it has no eigenstates.
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.
 
  • #16
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.
 

1. What is an Eigenstate in relation to Schrodinger Equation?

An eigenstate, also known as a stationary state, is a state in which a quantum system remains in a fixed state without any changes. In the context of Schrodinger equation, an eigenstate is a solution to the time-independent Schrodinger equation, which describes the energy levels and corresponding wavefunctions of a quantum system.

2. How does an eigenstate relate to the Schrodinger equation?

An eigenstate is a solution to the Schrodinger equation, which is a fundamental equation in quantum mechanics that describes the behavior of quantum systems. The eigenstates of a system are the possible states that the system can occupy, while the Schrodinger equation determines the probabilities of these states.

3. Can you give an example of an eigenstate in relation to the Schrodinger equation?

An example of an eigenstate in relation to the Schrodinger equation is the energy eigenstate of a particle in a one-dimensional box. In this case, the eigenstates are the standing wave solutions that correspond to different energy levels of the particle.

4. What is the significance of eigenstates in quantum mechanics?

Eigenstates are important in quantum mechanics because they represent the allowed states that a quantum system can occupy. The probabilities of transitioning between these states are determined by the Schrodinger equation, making eigenstates crucial in understanding the behavior of quantum systems.

5. How are eigenstates observed experimentally in quantum systems?

Eigenstates cannot be directly observed, but their effects can be observed through measurements and experiments. When a measurement is made on a quantum system, the system is forced into one of its eigenstates, and the corresponding eigenvalue is observed as the result of the measurement. These measurements can be repeated multiple times to determine the probabilities of the system being in each eigenstate.

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