Normalizing a wave function and finding probability density

In summary, the conversation discusses a homework problem involving a state of a particle bounded by infinite potential walls. The solution includes normalizing the wave function, calculating the probability of finding the particle between certain points, and finding the expectation value of the Hamiltonian operator. The conversation also includes a discussion on notation and terminology.
  • #1
Emspak
243
1

Homework Statement



A state of a particle bounded by infinite potential walls at x=0 and x=L is described by a wave function [itex]\psi = 1\phi_1 + 2\phi_2 [/itex] where [itex]\phi_i[/itex] 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 [itex]\langle \hat{H} \rangle[/itex]

Homework Equations



I wasn't sure Schrödinger's equation was necessary here: [itex]i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi(x,t)[/itex]
but I did remember that the solution to any wavefunction -- at least the stationary state -- is going to be [itex]\psi(x) = A sin(\frac{n\pi x}{a})[/itex] 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.

[itex]\psi = 1\phi_1 + 2\phi_2 [/itex]
[itex]\psi = (1\phi_1 + 2\phi_2)(1\phi_1^* + 2\phi_2^*)[/itex]

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

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

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

[tex]\int^L_0 |\psi|^2 dx = 1 \rightarrow \int^L_0 |5|^2 dx = 1 \rightarrow 25x = 1[/tex]

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

[itex]\psi = \frac{1}{5}\phi_1 + \frac{2}{5}\phi_2 [/itex]

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 [itex]\langle \hat{H} \rangle[/itex] - 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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  • #2
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.
 
  • #3
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?
 

Related to Normalizing a wave function and finding probability density

1. What is the purpose of normalizing a wave function?

The purpose of normalizing a wave function is to ensure that the total probability of finding the particle in any location is equal to 1. This allows us to interpret the wave function as a probability density function.

2. How do you normalize a wave function?

To normalize a wave function, we calculate the integral of the wave function squared over all space. Then, we divide the original wave function by the square root of that integral.

3. What does the probability density tell us about a particle's location?

The probability density, represented by the squared magnitude of the wave function, tells us the likelihood of finding the particle at a specific location in space. The higher the probability density, the more likely it is to find the particle in that location.

4. How does normalizing a wave function affect its shape?

Normalizing a wave function does not change its shape, but rather adjusts its amplitude to ensure that the total probability is equal to 1. This means that the shape of the wave function will remain the same, but it may appear taller or shorter depending on the original amplitude.

5. Can a non-normalized wave function still give accurate predictions?

No, a non-normalized wave function cannot give accurate predictions. This is because the total probability will not equal 1, which violates the laws of probability. Normalizing the wave function is necessary in order to obtain accurate predictions about the behavior of a particle.

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