How to show a wave function satisfies the shrodinger eqn

[richyw]

Homework Statement

If a state is a linear combination of two energy eigenstates, does its wavefunction satisfy the time independent shrodinger equation? I guess in general, how would I show if the wavefunctions of a state does or does not satisfy the time-independent schrodinger equation?

Homework Equations

[tex]H\left| E \right\rangle=E\left|E\right\rangle[/tex]
[tex]\psi_\alpha(x)=\left\langle x |\alpha\right\rangle[/tex]

The Attempt at a Solution

Honestly have no idea what to do.

 

[BruceW]

to start with, what is the time-independent Schrodinger equation?

 

[richyw]

to start with, what is the time-independent Schrodinger equation?

[tex]H\left| E \right\rangle=E\left|E\right\rangle[/tex]

 

[cepheid]

Homework Statement

If a state is a linear combination of two energy eigenstates, does its wavefunction satisfy the time independent shrodinger equation? I guess in general, how would I show if the wavefunctions of a state does or does not satisfy the time-independent schrodinger equation?

Homework Equations

[tex]H\left| E \right\rangle=E\left|E\right\rangle[/tex]
[tex]\psi_\alpha(x)=\left\langle x |\alpha\right\rangle[/tex]

The Attempt at a Solution

Honestly have no idea what to do.

Okay, first of all, by analogy with eigenvectors/values, an energy eigenstate is just a state (wavefunction) for which this is true: when you apply the Hamiltonian operator to it, you just get a constant (the eigenvalue, E) multiplied by it. So, if |E> is an energy eigenstate, then:$$H|E\rangle = E|E\rangle$$So energy eigenstates are states that satisfy the time-independent Schrodinger equation (which is the eigenvalue equation for the Hamiltonian).

Now, suppose we have two such eigenstates ##|E_1\rangle## and ##|E_2\rangle##. Since they are energy eigenstates, they must each separately satisfy the eigenvalue equation. So: ##H|E_1\rangle = E|E_1\rangle## AND ##H|E_2\rangle = E|E_2\rangle##

Now suppose we have a more general state ##|\psi\rangle## that is a linear combination of ##|E_1\rangle## and ##|E_2\rangle##:$$|\psi\rangle = \textrm{some combination of}~|E_1\rangle~\textrm{and}~|E_2\rangle$$What does it mean for psi to be a linear combination of the two eigenstates? I.e. what is a linear combination? I.e. can you fill in the right-hand side of the above equation?

Based on your answer to the above, what happens when you apply the Hamiltonian operator to psi? ##H|\psi\rangle =~?##

 

[richyw]

What does it mean for psi to be a linear combination of the two eigenstates? I.e. what is a linear combination? I.e. can you fill in the right-hand side of the above equation?

ok if I have two states [itex]\left| a\right\rangle[/itex] and [itex]\left| b\right\rangle[/itex] and they are energy eigenstates with energies [itex]E_a[/itex] and [itex]E_b[/itex], then a general linear combination of these states would be [tex]\left| \psi \right \rangle = c_1\left| a\right\rangle+c_2\left| b\right\rangle[/tex]

 

[richyw]

so if I wanted to show that the wavefunction [itex]\phi(x)=\left\langle x|\psi\right\rangle[/itex] satisfied the SE, how would I do this, and how would I find its eigenvalue?

 

[BruceW]

(in response to post 5) yep. so does psi satisfy the time-independent Schrodinger equation? i.e. what happens when you use the Hamiltonian operator on psi?

 

[cepheid]

ok if I have two states [itex]\left| a\right\rangle[/itex] and [itex]\left| b\right\rangle[/itex] and they are energy eigenstates with energies [itex]E_a[/itex] and [itex]E_b[/itex], then a general linear combination of these states would be [tex]\left| \psi \right \rangle = c_1\left| a\right\rangle+c_2\left| b\right\rangle[/tex]

Yeah, that's right. That is a linear combination. So what happens when you apply H to this state? Remember that H is a linear operator.

 

[richyw]

[tex]H\left| \psi \right \rangle = H\left( c_1\left| a\right\rangle+c_2\left| b\right\rangle\right)[/tex]
[tex]H\left| \psi \right \rangle = c_1 H\left| a\right\rangle+c_2H\left| b\right\rangle[/tex]
[tex]H\left| \psi \right \rangle = c_1 E_1\left| a\right\rangle+c_2E_2\left| b\right\rangle[/tex]

 

[richyw]

the part that is really messing me up is the "wave equation" part.

 

[BruceW]

yep nice. so is this the same as the time-independent Schrodinger equation for the state psi?

 

[richyw]

I guess?

 

[BruceW]

you said the time-independent Schrodinger is ##H|E\rangle = E|E\rangle## and for your situation, you have the state psi, so does psi fit this equation?

 

[richyw]

I have no idea.

 

[cepheid]

I have no idea.

What BruceW is saying is that for psi to satisfy the time-independent Schrodinger equation, it would have to be true that ##H|\psi\rangle = E|\psi\rangle## where E is *some constant.* Look at your above result for ##H|\psi\rangle##. Is it equal to "some constant" multiplied by psi?

 

[richyw]

What BruceW is saying is that for psi to satisfy the time-independent Schrodinger equation, it would have to be true that ##H|\psi\rangle = E|\psi\rangle## where E is *some constant.* Look at your above result for ##H|\psi\rangle##. Is it equal to "some constant" multiplied by psi?

no? I don't think it is?

 

[cepheid]

no? I don't think it is?

The answer is indeed no. But why aren't you sure about your answer? This is math after all, there is a definite answer. So don't guess, do the algebra. Take your psi and multiply it by some constant C (or E, or whatever). Is this the same as what you got for H|psi>?

 

[richyw]

are there any cases where it would be though? like what if one of the constants is imaginary, or if the two constants are equal?

 

[cepheid]

Use a more physical argument to motivate your answer: eigenstates (of the Hamiltonian) are states of definite energy. So if the particle is in an eigenstate, a measurement of its energy is sure to yield the same answer always. But if the particle is in a linear combination of two eigenstates, can this wavefunction psi tell you with certainty what the energy will be, or can it only tell you about the probability of measuring a given value? Based on that, would you expect the linear combination of two eigenstates to also be an eigenstate, or not?

 

[cepheid]

are there any cases where it would be though? like what if one of the constants is imaginary, or if the two constants are equal?

If the constants are equal (i.e. two distinct states have the same value -- they are said to be degenerate states), then yes, it would work. Their linear combination would also be an eigenstate with that same energy.