What is the Solution to the One-Dimensional Particle in an Energy Well Problem?

[Dick]

Sure. So phi*=phi.

 

[TFM]

okay so:


\phi = \phi^* = Asin(\frac{n\pi}{L}x)

Asin(\frac{n\pi}{L}x)*Asin(\frac{n\pi}{L}x) = 1

= A^2 sin^2(\frac{n\pi}{L}x) = 1

= A^2 = \frac{1}{sin^2(\frac{n\pi}{L}x)}

= A = \sqrt{\frac{1}{sin^2(\frac{n\pi}{L}x)}}

This doesn't seem right to me though...?

 

[Dick]

Integrate. Integrate. Integrate. I've mentioned that several times. Look up the definition of <> and normalization. You have to integrate x from 0 to L on the third line.

 

[TFM]

Sorry, so:

\int{\phi^* \phi}dx = 1

so:

\int{Asin(\frac{n\pi}{L}x)*Asin(\frac{n\pi}{L}x)} dx = 1

A^2\int{sin(\frac{n\pi}{L}x)*sin(\frac{n\pi}{L}x)} dx = 1

A^2\int{sin^2(\frac{n\pi}{L}x)dx = 1

A^2[\frac{n\pi}{2L}x - \frac{n\pi}{4L}sin(\frac{2n\pi}{L}x)]^L_0 = 1

A^2[\frac{n\pi}{2L}0 - \frac{n\pi}{4L}sin(\frac{2n\pi}{L}(0)) - \frac{n\pi}{2L}(L) - \frac{n\pi}{4L}sin(\frac{2n\pi}{L}L)] = 1

A^2[(\frac{n\pi}{2}] = 1

A^2 n\pi = 2

A^2 = \frac{2}{n\pi}

A = \sqrt{\frac{2}{n\pi}}

Does tgis look better?

 

[Dick]

Well, I get the integral to be L/2. No pi's, no n's. Better check the integration. Use sin^2(kx)=(1-cos(2kx))/2. The cos part doesn't contribute anything. It averages out. How did you get all those n*pi/L parts outside?

 

[TFM]

Okay, so:

A^2\int{sin^2(\frac{n\pi}{L}x)dx = 1

Using:

sin^2 (kx) = \frac{1}{2} - \frac{cos(kx)}{2}

Gives:

A^2\int{\frac{1}{2} - \frac{cos(kx)}{2}}dx = 1

A^2[{\frac{x}{2} - \frac{k}{2} sin(kx)}]^L_0 = 1

A^2[{\frac{L}{2} - \frac{k}{2} sin(kL)}] = 1

k = n\pi/L

A^2[{\frac{L}{2} - \frac{n\pi}{2} sin(n\pi)}] = 1

sin of n*pi = 0

A^2[{\frac{L}{2}] = 1

A^2{\frac{L}{2} = 1

A^2 = {\frac{2}{L}

gives A to be:

A = \sqrt{\frac{2}{L}}

Is this better now?

 

[TFM]

OKay, I have looked in the book and it does appear to be the right answer - should I have done this part in the previous section?

So: \sqrt{\frac{2}{L}}sin(\frac{n\pi}{L}x)

Okay, so now:

\left\langle \hat{p} \right\rangle = \int\phi^*\hat{p}\phi

since we have found that phi* = phi, I assume:

\left\langle \hat{p} \right\rangle = \int^L_0\sqrt{\frac{2}{L}}sin(\frac{n\pi}{L}x) (-i\hbar \frac{\partial}{\partial x}\sqrt{\frac{2}{L}}sin(\frac{n\pi}{L}x)dx

and thus:

\left\langle \hat{p} \right\rangle = \int^L_0\sqrt{\frac{2}{L}}sin(\frac{n\pi}{L}x) (-i\hbar \sqrt{\frac{2}{L}}\frac{\partial}{\partial x}sin(\frac{n\pi}{L}x)dx

and:

\left\langle \hat{p} \right\rangle = \int^L_0\sqrt{\frac{2}{L}}sin(\frac{n\pi}{L}x) (-i\hbar \sqrt{\frac{2}{L}}\frac{n\pi}{L} cos(\frac{n\pi}{L}x)dx

Rearrange:

=-\frac{2n\pi}{L^2}\int^L_0 sin(\frac{n\pi}{L}x)(cos(\frac{n\pi}{L}x)

Does this look okay so far?

 

[TFM]

Okay, I am not sure for the p-hat one I just did, but I have done the x ones, and I have got:

&lt;\hat{x}&gt; = \phi^*x\phi

= \frac{2}{L}\int^L_0 sin (\frac{n\pi}{L}x)xsin (\frac{n\pi}{L}x)

= \frac{2}{L}\int^L_0 x sin^2 (\frac{n\pi}{L}x)

using the substituion for sin^2:

= \frac{2}{L}\int^L_0 \frac{x}{2} x\frac{cos(2kx){2}

using the product rule, I get:

\frac{2}{L}[\frac{xksin(2kx)}{2} + \frac{x^2}{2}\frac{cos2kx}{2}]

insert values:

\frac{2}{L}[\frac{L ksin(2kL)}{2} + \frac{L^2}{2}\frac{cos2kL}{2}]

insert k:

\frac{2}{L}[\frac{L \frac{n\pi}{L}sin(2\frac{n\pi}{L}L)}{2} + \frac{L^2}{2}\frac{cos2\frac{n\pi}{L}L}{2}]

Cancel down:

\frac{2}{L}[\frac{n\pi sin(2n\pi)}{2} + \frac{L^2}{2}\frac{cos2n\pi}{2}]

Since sin n*pi = 0, cos npi = 1:

\frac{2}{L}[\frac{L^2}{2}\frac{1}{2}]

and:

\frac{2L^2}{2L}\frac{1}{2}

Giving:

&lt;\hat{x}&gt; = \frac{L}{2}

And, similarly for <x^2>

&lt;\hat{x}&gt; = \phi^*x^2\phi

= \frac{2}{L}\int^L_0 sin (\frac{n\pi}{L}x)x^2 sin (\frac{n\pi}{L}x)

= \frac{2}{L}\int^L_0 x^2 sin^2 (\frac{n\pi}{L}x)

using the substituion for sin^2:

= \frac{2}{L}\int^L_0 \frac{x^2}{2} x^2\frac{cos(2kx){2}

using the product rule, I get:

\frac{2}{L}[\frac{x^2ksin(2kx)}{2} - \frac{x^3}{6}\frac{cos2kx}{2}]

inserting the values for k and x:

\frac{2}{L}[\frac{L^2ksin(2kL)}{2} - \frac{L^3}{6}\frac{cos2kL}{2}]

\frac{2}{L}[\frac{L^2\frac{n\pi}{L}sin(2\frac{n\pi}{L}L)}{2} - \frac{L^3}{6}\frac{cos2\frac{n\pi}{L}L}{2}]

Cancel down:

\frac{2}{L}[\frac{L n\pi sin(2n\pi}{2} - \frac{L^3}{6}\frac{cos2n\pi}{2}]

Using same cos and sin rules:

\frac{2}{L}[- \frac{L^3}{6}\frac{1}{2}]

-\frac{L^2}{6}

Thus:

\Delta \hat{x} = \sqrt{&lt;\hat{x^2}&gt; - &lt;\hat{x}&gt;^2}

and:

\Delta \hat{x} = \sqrt{&lt;-\frac{L^2}{6}&gt; - &lt;{\frac{L^2}{4}}&gt;^2}

\Delta \hat{x} = \sqrt{&lt;-\frac{L^2}{6}&gt; - &lt;{\frac{L^2}{4}}&gt;^2}

Does this look okay?

 

[Dick]

For <p> you've got the right general form, but I'm wondering about some of the constants (not that they'll matter in the end). <x> the answer came out ok, the stuff in between doesn't look completely ok. <x^2> CAN'T be negative. Why not? Think about it.

 

[TFM]

Okay,

<x^2> can't be negatvive because it is squared,;

I was writing the working out down to hand in, but I didn't use copy from my notes, and it looked slightly better, but it gave me:

<x> = L

<x^2> = 2/3 L^2

trouble is when working out the Delta:

\Delta\hat{x} = \sqrt{&lt;\hat^2&gt; - &lt;\hat&gt;^2}

trouble is I get a negative equation under the sqare root, which I know shouldn't be...?

 

[Dick]

If you get a negative under the square root, then you made a mistake. I don't get <x>=L, I get <x>=L/2. <x^2> doesn't seem to come out as 2L^2/3 either. These are just integration problems now, not QM. So it doesn't take any deep thinking. You just have to do the integrations carefully.