Wave Packet and Standard Deviations

[erok81]

Homework Statement

A free particle has wave function (i have no idea how to get these come out correctly, so I can't use LaTeX.

Edit on the LaTeX. I'm getting closer! Ok...I can't get rid of the space but that's another subject.

\Psi(x)=\left\{\begin{array}{cc}N(1+\frac{x}{a}),&amp;\mbox{ if }<br /> -a \leq x &lt; 0\\N(1-\frac{x}{a}),&amp;\mbox{ if } 0 \leq x &lt; a \end{array}\right

Zero elsewhere.

Where a is a parameter and N is a normalization constant.

a) Find N
b) Compute <x>, <x²> and find the standard deviation.

Homework Equations

The Attempt at a Solution

I think I can find the second part of this question but I am not sure about N. I can't find anything in my book or anything we've gone over in class. I've tried google but can't find anything useful.

So first off...any ideas on this N?

 

[cepheid]

I think I can find the second part of this question but I am not sure about N. I can't find anything in my book or anything we've gone over in class. I've tried google but can't find anything useful.

So first off...any ideas on this N?

Remember, the probabilistic interpretation of the wavefunction says that the probability of finding the particle somewhere between position x₁ and position x₂ is given by:

\int_{x_1}^{x_2} |\Psi(x)|^2\,dx

Now, the particle is certain to be somewhere. So, what is the probability of finding the particle anywhere in the entire possible range of positions where it can be?\int_{-\infty}^{+\infty} |\Psi(x)|^2\,dx = ~ ?

The answer to this question is the reason for the so-called normalization condition on the wavefunction.

 

[erok81]

Oh what do you know...that integral makes sense now.

This is an exremely stupid question but I cannot remember how to solve this the rest of the way. So I did the first interval and came up with this.

<br /> \int_{-\infty}^{+\infty} |\Psi(x)|^2\,dx = \left[ \frac{aN^{2}}{3}\left(1+\frac{x}{a}\right)^3\left|_{-\infty}^{+\infty} \right]<br />

I can't figure out the final answer for this. Well I tried and came up with zero. Am I still doing this wrong?

 

[cepheid]

You didn't answer my question! The answer to my question was that the integral over all space (i.e. from -∞ to +∞) has to be equal to 1, because it represents the probability that a measurement of the particle's position will find it to be somewhere in the entire range of possible positions. This condition (that the integral over all space must equal 1), is called the normalization condition. It is a condition that any position wavefunction has to satisfy. Setting the constant 'N' to the right value allows you to satisfy the condition and normalize the wavefunction. That's why N is called the "normalization constant" in your original problem statement. So, set the integral equal to 1, and solve for N.

As for the mechanics of actually solving the integral: I think you're going a little off track here, so let me get you started from the beginning. The wavefunction is ZERO anywhere outside the interval [-a,a]. So the integral reduces to:

\int_{-a}^{+a} |\Psi(x)|^2\,dx = 1

Since Ψ is a piecewise function, this has to be split into two separate integrals, the first of which is:

\int_{-a}^{0} N^2\left(1+\frac{x}{a}\right)^2\,dx =N^2 \int_{-a}^{0} \left(1+\frac{x}{a}\right)\left(1+\frac{x}{a}\right)\,dx

= N^2 \int_{-a}^{0} \left(1+2\frac{x}{a} + \frac{x^2}{a^2}\right)\,dx

The integral of a sum of functions is the sum of the integrals of those individual functions. Therefore, this splits into three separate integrals, all of which are manageable. Can you take it from here?

 

[erok81]

Ah! That is what I was missing. I originally had my integrals set up similar to what you have but didn't know what to do after that.

Thanks for detail on why that is set to one. I went back and looked in my text and still didn't find it, so I would never have gotten this one. I've got class and work for the next 14 hours but after that I'll give it another try and post back.

Thanks again for the help thus far.

 

[erok81]

Ok...here we go.

<br /> = N^2 \int_{-a}^{0} \left(1+2\frac{x}{a} + \frac{x^2}{a^2}\right)\,dx = \frac{1}{3}N^2 a<br />

Equation two also gives the same result.

Solving for N:

N= \pm\sqrt{\frac{3}{a}}

Thanks again for the direction. I probably wouldn't have figured that out alone.

Now onto the next part. Shouldn't be too hard as it's just.

\left\langle f(x) \right\rangle=\int \psi (x) f(x) \psi^* (x) dx

I might post again on the third part since I am not 100% sure.

Estimate the momentum spread and the kinetic energy of the particle.

 

[cepheid]

Ok...here we go.

<br /> = N^2 \int_{-a}^{0} \left(1+2\frac{x}{a} + \frac{x^2}{a^2}\right)\,dx = \frac{1}{3}N^2 a<br />

Equation two also gives the same result.

Agreed.

Solving for N:

N= \pm\sqrt{\frac{3}{a}}

Not agreed. If both integrals are equal to \frac{1}{3}N^2 a, then what is the total result for the original integral that you were trying to compute?

Thanks again for the direction. I probably wouldn't have figured that out alone.

Now onto the next part. Shouldn't be too hard as it's just.

\left\langle f(x) \right\rangle=\int \psi (x) f(x) \psi^* (x) dx

I might post again on the third part since I am not 100% sure.

Estimate the momentum spread and the kinetic energy of the particle.

Yes, feel free to post again if you need help with subsequent parts.

 

[erok81]

Not agreed. If both integrals are equal to \frac{1}{3}N^2 a, then what is the total result for the original integral that you were trying to compute?

Yep. I see now. That is actually how I was doing it at first. But stupidly assumed I only needed to find it once (since it was the same) and apply it to each. Not add them.

So rephrased.

<br /> N= \pm\sqrt{\frac{3}{2a}}<br />

The rest wasn't too bad.

Thanks again for the help. Beyond appreciated.