# How to show a wave function satisfies the shrodinger eqn

1. Dec 9, 2013

### richyw

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

$$H\left| E \right\rangle=E\left|E\right\rangle$$
$$\psi_\alpha(x)=\left\langle x |\alpha\right\rangle$$

3. The attempt at a solution

Honestly have no idea what to do.

2. Dec 9, 2013

3. Dec 9, 2013

### richyw

$$H\left| E \right\rangle=E\left|E\right\rangle$$

4. Dec 9, 2013

### cepheid

Staff Emeritus
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? $H|\psi\rangle =~?$

5. Dec 9, 2013

### richyw

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

6. Dec 9, 2013

### richyw

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

7. Dec 9, 2013

### 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?

8. Dec 9, 2013

### cepheid

Staff Emeritus
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.

9. Dec 9, 2013

### richyw

$$H\left| \psi \right \rangle = H\left( c_1\left| a\right\rangle+c_2\left| b\right\rangle\right)$$
$$H\left| \psi \right \rangle = c_1 H\left| a\right\rangle+c_2H\left| b\right\rangle$$
$$H\left| \psi \right \rangle = c_1 E_1\left| a\right\rangle+c_2E_2\left| b\right\rangle$$

10. Dec 9, 2013

### richyw

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

11. Dec 9, 2013

### BruceW

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

12. Dec 9, 2013

### richyw

I guess?

13. Dec 9, 2013

### 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?

14. Dec 9, 2013

### richyw

I have no idea.

15. Dec 9, 2013

### cepheid

Staff Emeritus
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?

16. Dec 9, 2013

### richyw

no? I don't think it is?

17. Dec 9, 2013

### cepheid

Staff Emeritus
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>?

18. Dec 9, 2013

### 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?

19. Dec 9, 2013

### cepheid

Staff Emeritus
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?

20. Dec 9, 2013

### cepheid

Staff Emeritus
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.