Expectation Value/ Evaluating Gaussian Integral

[atomicpedals]

Homework Statement

I'm re-hashing a problem from my notes; given the wave function

$$\psi(x)=Ne^{-(x-x_0)/2k^2}$$

Find the expectation value <x>.

Homework Equations

The normalization constant N for this is in my notes as $$N^2=1/\sqrt{2 \pi k^2}$$ $$N=1/(2\pi k^2)^{(1/4)}$$ It should be solved through a u substitution.

The Attempt at a Solution

$$<x>=\int \psi(x)^{*}x \psi(x) dx=\int x |\psi(x)|^{2} dx$$
$$=\int x N^2 e^{-(x-x_0)^2/k^2} dx$$
$$=N^2 \int x e^{-(x-x_0)^2/k^2} dx$$
$$u=x-x_0, du=x dx$$
$$<x>=N^2 \int e^{-u^2/k^2}du$$
This is where I stumble. Assuming I wrote the problem down correctly the expectation value should come out to <x>=x₀. But I'm just not seeing it which leads me to believe that I'm not solving the final integral correctly.

 

[dacruick]

N^2?

 

[atomicpedals]

Woops, type-o on my part. I've edited the original post...the question should be correct now.

 

[dacruick]

$$<x>=N^2 \int e^{-u^2/k^2}du$$

where did that x go? you subbed u = x - xo but you didnt set x = to u + xo

 

[D H]

$$u=x-x_0, du=x dx$$

This is obviously wrong.

Try using u=(x-x₀)², or use the substitution you used (but do it correctly).

 

[atomicpedals]

If I set $$u=x-x_0$$ and say $$x=u+x_0$$ I then get $$xdx=du$$ don't I?

 

[atomicpedals]

Yeah, moment of clarity here... let me re-do it with $$u=(x-x_0)^2$$.

 

[dacruick]

If I set $$u=x-x_0$$ and say $$x=u+x_0$$ I then get $$xdx=du$$ don't I?

no, just du = dx

 

[atomicpedals]

So in a fit of mathematical confusion I decided to take the problem all the way back to the beginning. And I might be closer to having done it properly...I hope.

Given my wave function $$\psi(x)=Ne^{-(x-x_0)/2k^{2}}$$ I needed to determine the normalization constant N which should be found via $$\int|\psi(x)|^{2}dx=1$$ so I found

$$\int|Ne^{-(x-x_0)/2k^{2}}|^{2}dx=1$$
$$\int|N^{2}e^{-(x-x_0)^2/k^2}|dx=N^{2}\int e^{-(x-x_0)^2/k^2}dx$$
My substitution was $$u=\frac{x-x_0}{k}, kdu=dx$$ and then
$$N^{2}\int ke^{-u^2}du=N^{2}k\int e^{-u^2}du=N^{2}k\sqrt{pi}=1$$
$$N=\frac{1}{\pi^{1/4}\sqrt{k}}$$ and $$N^2=\frac{1}{\sqrt{\pi}k}$$
Similarly then, the expectation value <x> is
$$<x>=\int \psi(x)^{*}x \psi(x) dx=\int x |\psi(x)|^{2} dx$$
$$=\int x N^{2} e^{-(x-x_0)^2/k^{2}} dx$$
$$=N^{2} \int x e^{-(x-x_0)^2/k^{2}} dx$$
$$u=\frac{x-x_0}{k}, kdu=dx$$
$$<x>=N^{2} \int xk e^{-u^{2}}du$$
$$<x>=N^{2} xk \int e^{-u^{2}}du=N^{2} xk \sqrt{pi}=\frac{kx\sqrt{\pi}}{\sqrt{\pi}k}=x$$

Success?

 

[vela]

No, you made several errors. You can't pull x out of the integral because x is a function of u.

What's the initial wave function? In one step, you essentially said (e^x)² = e^x2, but that's incorrect since (e^x)² = e^2x.

 

[atomicpedals]

It's clearly been a long time since I've done much with u-subs. How would I handle the x in the integrand? (Though that may be immaterial to the greater error with the exponent)

The wave function
$$\psi(x)=Ne^{-(x-x_0)/2k^{2}}$$
$$\psi(x)^{2}=N^{2}e^{-(x-x_0)/k^{2}}$$
This then brings me back to my original stumbling point of choosing an appropriate substitution.

 

[vela]

The problem with that wave function is that it diverges as x goes to negative infinity. It's not normalizable. Are you sure you don't mean
$$\psi(x) = Ne^{-(x-x_0)^2/2k^2}$$
If so, after you make the substitution, you find the normalization constant the way you did above.

When finding <x>, you do the substitution, but you have to now write x in terms of u.

 

[atomicpedals]

Oh, wow, yes. I did mean $$\psi(x) = Ne^{-(x-x_0)^2/2k^2}$$ (clearly I've been looking at this too long) This would then square to $$\psi(x)^2 = N^2e^{-(x-x_0)^2/k^2}$$. Is the normalization constant N I found in the first part correct (success despite myself...although I'm suspecting not)?

 

[vela]

Yes, the two mistakes canceled each other out.

 

[atomicpedals]

Now that's refreshing, so wrong I was right!

Ok, going back to the expectation value then; I have the dangling x in the integrand post substitution $$N^{2}k\int x e^{-u^{2}}du$$ how do I handle the x? Solve for x in terms of u? And since it's been about 7 years since I took Calc II I may need the basic lesson here.

 

[wizang]

Try setting x in terms of u. I didn't follow the whole thread, but I believe u=(x-xo)/k. So solve for x in terms of u, put that in and the integral should be more manageable... though you'll still need integration by parts, no way around that.

 

[atomicpedals]

Right on,
$$u=\frac{x-x_0}{k}$$
$$x=ku+x_0$$
And so,
$$N^{2}k\int ku+x_0 e^{-u^{2}}du$$

 

[vela]

Use parentheses!
$$N^2 k\int_{-\infty}^\infty (ku+x_0) e^{-u^{2}}\,du = N^2 k^2\int_{-\infty}^\infty ue^{-u^{2}}\,du + N^2 kx_0\int_{-\infty}^\infty e^{-u^{2}}\,du$$Can you see by inspection why the first integral is equal to 0?

 

[atomicpedals]

Oh wow, that is fantastically elegant!