Normalizing a wave function and finding probability density

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The discussion revolves around normalizing a wave function for a particle in an infinite potential well and calculating the probability density and expectation value of the Hamiltonian. The wave function is expressed as a superposition of stationary states, and the user attempts to normalize it by integrating the squared wave function over the interval from 0 to L. There is confusion regarding the normalization constant and the integration process, with the user mistakenly concluding that the normalization factor is 5 instead of the correct value. Additionally, there is uncertainty about calculating the expectation value of the Hamiltonian operator and its notation. Clarification on these points is sought to enhance understanding of the concepts involved.
Emspak
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Homework Statement



A state of a particle bounded by infinite potential walls at x=0 and x=L is described by a wave function \psi = 1\phi_1 + 2\phi_2 where \phi_i are the stationary states.
a) Normalize the wave function.
b) What is the probability to find the particle between x=L/4 and x=3L/4?
c) Calculate the expectation value of the Hamiltonian operator \langle \hat{H} \rangle

Homework Equations



I wasn't sure Schrödinger's equation was necessary here: i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi(x,t)
but I did remember that the solution to any wavefunction -- at least the stationary state -- is going to be \psi(x) = A sin(\frac{n\pi x}{a}) where a is the length of your "box" for the particle. So in this case a=L.

The Attempt at a Solution



We have a situation where V(x) = 0 0 < x < L and V(x) = infinity outside of that. So the V(x) term for inside the well disappears (it's zero).

The probability of the particle being at any point from 0 to L is 1. So I need to integrate the wave functions squared over that interval. By the superposition principle it is OK to just add them.

\psi = 1\phi_1 + 2\phi_2
\psi = (1\phi_1 + 2\phi_2)(1\phi_1^* + 2\phi_2^*)

multiply this out
\psi = (1\phi_1^* \phi_1 + 2\phi_1 \phi_2^* + 2\phi_1^* \phi_2 + 4\phi_2^*\phi_2)

SInce the phi functions are eigenvalues, the ones on the diagonal of the matrix are the only ones not zero. So we get
\psi = (1\phi_1^* \phi_1 + 4\phi_2^*\phi_2) = (1 + 4)

because the complex conjugate of a function multiplied by a function is 1.

That makes the whole thing add up to five. and since the probability of finding the particle on the interval 0 to x is

\int^L_0 |\psi|^2 dx = 1 \rightarrow \int^L_0 |5|^2 dx = 1 \rightarrow 25x = 1

so x = 1/5 for the whole interval, (since that is the square root of 1/25) so normalizing the wave function I should get

\psi = \frac{1}{5}\phi_1 + \frac{2}{5}\phi_2

and for the probability that the particle is at L/4 and 3/4 L

(25L/4) and (75L/4)

Now, if someone could tell me where I am getting lot and doing this completely wrong :-)

Actually I know this is wrong, because the probabilities should add up to one, at least with the stationary states.

After that I get even more confused. I think -- and I stress think -- I have some vague idea of how to get expectation values, but whenever someone mentions stuff like "Hamiltonian operator" I want to run away and hide. :-) More seriously, I am trying to grasp what exactly is meant by the notation \langle \hat{H} \rangle - some of it is a notational question, but i get confused because I am never sure if they want the Hamiltonian like what you do in mechanics or something else. I feel like if someone could explain that I'd be a lot further along.

thanks in advance.

PS I put this up before in advanced HW, but perhaps it goes here. (Nobody answered, I might have put up the wrong tags) The line between what counts as advanced around here and what is basic seems rather fluid :-) But since I am clearly the dumbest guy in the class I'll start here. :-)
 
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You have the right idea, but you're making some silly mistakes.
withe ψ = N (ψ1 + 2 ψ2) and integrate ∫|ψ|2 dx from zero to L and set the integral to 1 to find the normalization factor N. Tip: N is not equal to 5.
 
I did that tho, didn't I? I got x=1/25. I'm not sure I understand what you are referring to.

When I do the integration I get the ∫|ψ|2 dx = ∫|5|2 dx. Is that not correct? Because if you integrate that from 0 to L you would get 25* (x) from 0 to L, or 25L = 1. That would make L = 1/25, no?

And in that case my N would be 1/25? Is that what you are saying?
 
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