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:

$$<\hat{x}> = \phi^*x\phi$$

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

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

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:

$$<\hat{x}> = \frac{L}{2}$$

And, similarly for <x^2>

$$<\hat{x}> = \phi^*x^2\phi$$

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

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

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:

$$\Delta \hat{x} = \sqrt{<\hat{x^2}> - <\hat{x}>^2}$$

and:

$$\Delta \hat{x} = \sqrt{<-\frac{L^2}{6}> - <{\frac{L^2}{4}}>^2}$$

$$\Delta \hat{x} = \sqrt{<-\frac{L^2}{6}> - <{\frac{L^2}{4}}>^2}$$

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:

$$\Delta\hat{x} = \sqrt{<\hat^2> - <\hat>^2}$$

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.