What is the significance of the ooze wave function and its normalization?

[UrbanXrisis]

The following is a "ooze" wafve function:

$$\Psi_{ooze} (x,t)=\frac{1}{K} \left( \Psi_1 + \Psi_2+...+\Psi_{1000} \right)$$


1. I am to find the value of K, but I don't even know what it represents. Is K the coefficient to normalize the coefficient to 1?

2. Probability where energy $$E_q$$ can be observed?


$$E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$$

$$\sum_{n=1}^q \frac{n^2 \pi^2 \hbar^2}{2mL^2}$$

$$\frac{ \pi^2 \hbar^2}{2mL^2} \sum_{n=1}^q n^2$$

3. Then I am to find the average energy predicted to be observed. I am not sure what this even means. I am guessing: $$E_{avg} = \frac{n^2 \pi^2 \hbar^2}{2nmL^2}$$ ??

 

[dextercioby]

I don't know what you're doing at 2, but for 1., yes, it's a normalization factor.

Daniel.

 

[Galileo]

1) A constant factor in front the wavefunction is always for normalization. It can also have a phase, but that's of no physical significance.

2) Ok, from the energies I see this is a square potential well problem. Why are you summing all the energies from $E_1$ to $E_q$? (And what does it mean?)
You have to find the probability (which is a number between 0 and 1) to measure $E_q$ when you do a measurement of the energy. So what's the general rule for finding the probability of getting a certain result?

3. If you find the probabilities in 2) then I`m sure you can calculate the average.

 

[UrbanXrisis]

I know how to find the probability for a wavefunction, but not the probability for energy. For a wavefunction, you multiply the complex conugate of the wavefunction but what about energy? and how does the average energy relate to this?

 

[Galileo]

I know how to find the probability for a wavefunction, but not the probability for energy. For a wavefunction, you multiply the complex conugate of the wavefunction but what about energy? and how does the average energy relate to this?

What do you mean 'I know how to find the probability for a wavefunction'. What $|\Psi|^2$ tells you is the probability density for the position of the particle. The phrase 'probability for a wavefunction' doesn't make sense in this context.

The probability of measuring $E_n$ is:
$$P(E_n)=|\langle \Psi_n|\hat H|\Psi\rangle|^2$$
$$\langle \Psi_n|\hat H|\Psi\rangle=\int_{-\infty}^{\infty}\Psi_n^*(x,t) \hat H \Psi(x,t)dx$$
Surely something like that should be in your book?

Once you know the probabilities for all the energies, you can calculate the average (also called the expected value, which is misleading). But that's not physics per se, that's probability, or statistics:
$$\langle A\rangle=\sum_n a_nP(a_n)$$

 

[UrbanXrisis]

how would I find K? I am really stumped by this

 

[Galileo]

K should be such that the wavefunction is normalized.
I
So:
$$\int_{-\infty}^{+\infty}|\Psi(x)_{ooze}|^2dx=1$$

Now write out the integrand. You'll get a lot of terms, but note all the cross terms integrate to zero, since the $\Psi_i$ are orthogonal.

 

[UrbanXrisis]

$$\int \frac{1}{K}\frac{1}{K} \Psi_{1}* \Psi_{1} (x,t)+...+\frac{1}{K}\frac{1}{K} \Psi_{1000}* \Psi_{1000} (x,t)=1$$

$$= \frac{1000}{k^2} =1$$

$$k=\sqrt {1000}$$

something like that?

 

[Galileo]

Yep.

 

[UrbanXrisis]

so if I was to find the average energy it would be $$\frac{1}{1000} E_1 +\frac{1}{1000} E_2 +...+\frac{1}{1000} E_{1000}$$ ?

 

[Galileo]

Yes again.