# Wave function and infinite square well potential

## Homework Statement

An electron in a one-dimensional infinite square well potential of length L is in a
quantum superposition given by ψ = aψ1+bψ2, where ψ1 corresponds to the n = 1 state, ψ2 corresponds to the n = 2 state, and a and b are constants. (a) If a = 1/3, use the
normalization requirement for ψ to determine the value of b. (b) If we perform a
measurement of the energy of the electron, what is the probability we will measure E1?
What is the probability we will measure E2?

Don't know.

## The Attempt at a Solution

So basically I have to find the ψ of an electron in the ground state and then in the n = 2 state? Do I solve for that? And how do I normalize a wave function that doesn't have "i" in the exponent? I've only learned that normalizing is making the "i" a "-i" as the conjugate, and then multiplying it with the original function. How do I normalize something without "i" in its exponent?

Thanks.
And then

Simon Bridge
Homework Helper
So basically I have to find the ψ of an electron in the ground state and then in the n = 2 state?
No - that is not what it says.
The n states are stationary states - the problem is telling you that the particle is not in a stationary state. It is in a single non-stationary state that happens to have the same shape as a sum of the first two stationary states.

The stationary states are not the only solutions to the schodinger equation for the infinite square well ... any linear superposition of them are also solutions.

And how do I normalize a wave function that doesn't have "i" in the exponent? I've only learned that normalizing is making the "i" a "-i" as the conjugate, and then multiplying it with the original function. How do I normalize something without "i" in its exponent?
Exactly the same way:

Consider an arbitrary complex number z in terms of a,b which are real and positive:
##z=a+ib \Rightarrow z^\star = a-ib## which you are used to... but what if z=a alone?
##z=a \Rightarrow z=a+i0## now do it.
##z^\star = a-i0## but zero times i is zero so ##z^\star = a##

...thus: the complex conjugate of a real number is itself.

Tread carefully: it says to use the "normalization condition" - what does that mean for a and b?

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I don't understand what to use as ψ. Do I use ei(kx-wt)? That's a wave function I see in this book. It's confusing because whenever they need to do something to a wave function, they pull one out of thin air and operate on it.
I do need a wave function for the electron in order to do this problem, right? So how do I create one?

The normalization requirement is the "requirement" that the probability of the particle existing within the well is 100% right? So I set ∫ψ*ψ=1, plug in the wave function that I find for the electron, and then normalize it?

Thanks.

CAF123
Gold Member
I don't think you have to sub anything for ##\psi_1## and ##\psi_2##. The orthogonality of the ##\psi_i## will allow you to cancel terms.

So they cancel and I end up with 0?

CAF123
Gold Member
So they cancel and I end up with 0?
No, what does the orthogonality condition state? i.e what is ##\langle \psi_i | \psi_j \rangle = \int \psi_i^* \psi_j \,dx## equal to?

No, what does the orthogonality condition state? i.e what is ##\langle \psi_i | \psi_j \rangle = \int \psi_i^* \psi_j \,dx## equal to?

Oh, so the conjugate of a wave function times the wave function itself are orthogonal? Ok.
I believe that it's equal to 1, because 1 is the probability of the particle existing in the well. So orthogonality condition is the same as the probability?

CAF123
Gold Member
Oh, so the conjugate of a wave function times the wave function itself are orthogonal? Ok.
I believe that it's equal to 1, because 1 is the probability of the particle existing in the well.
That is correct if ##i = j##, but what happens if ##i \neq j##? Have you come across orthogonality in your course yet?
The physics is that the ##\psi_i## are orthogonal meaning they represent different physical states of the wave function. This means you cannot express each of the eigenstates as a linear combination of each other, but the wave function can be expressed as a linear combination of the eigenstates, which is precisely what you have in your problem.

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vela
Staff Emeritus
Homework Helper

## Homework Statement

An electron in a one-dimensional infinite square well potential of length L is in a
quantum superposition given by ψ = aψ1+bψ2, where ψ1 corresponds to the n = 1 state, ψ2 corresponds to the n = 2 state, and a and b are constants. (a) If a = 1/3, use the
normalization requirement for ψ to determine the value of b. (b) If we perform a
measurement of the energy of the electron, what is the probability we will measure E1?
What is the probability we will measure E2?

Don't know.

## The Attempt at a Solution

So basically I have to find the ψ of an electron in the ground state and then in the n = 2 state? Do I solve for that? And how do I normalize a wave function that doesn't have "i" in the exponent? I've only learned that normalizing is making the "i" a "-i" as the conjugate, and then multiplying it with the original function. How do I normalize something without "i" in its exponent?

Thanks.
And then
As CAF123 noted, you don't actually need expressions for the eigenstates to solve this problem. Nevertheless, you should know what they are. The infinite square well problem is a standard problem in intro quantum mechanics. It's probably worked out in your textbook.

Simon Bridge
Homework Helper
Lets not get ahead of ourselves - look at one thing at a time or it will get confusing.
(a) If a = 1/3, use the normalization requirement for ψ to determine the value of b.
... what is the "normalization requirement"?

if ##\psi = \sum_i c_1\psi_i##
if each ##psi_i## are normalized, then there is a requirement on the ##c_i##'s that make ##\psi## normalized as well.

That is the normalization requirement.

Why it works (that "orthogonality stuff) is another question - which should be covered in your text and your course notes along with the requirement itself. I want you to get used to using them. Meantime - when you see it you will also see what you are expected to do.

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