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

[TFM]

Homework Statement

Consider a particle that is confined in a one-dimensional box, ie in a potential

V(x) = 0 for 0 \leq x \leq L, or \infty when x < 0, x > L

(i) Determine the solutions Φn(x) of the stationary Schrödinger equation for this problem. Make sure that you have normalized them correctly.

(ii) Calculate the energy eigenvalue En corresponding to Φn(x).

(iii) Use the results of (i) and (ii) and write down the complete time-dependent wave function Ψn(x,t) for the nth stationary state in this potential.

( iv) For the nth stationary state calculate .

(v) Use the results of (iv) to check whether the Heisenberg Uncertainty relation is satisfied for the nth stationary state? Which state comes closest to the minimum uncertainty?

Homework Equations

The Attempt at a Solution

Okay, I have done (i), and have got:

k = \frac{n\pi}{L}

and

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

Where n is plus/minus 1,2,3...

Okay, but I am not sure on the second part.

I have a book, and it has:

E_n = \frac{\hbar^2k_n^2}{2m}

And I think this is the irght formula. The trouble is that IO believe that this equation isbasically the answer, just sub in the result I got above, but I don't know where the equation has come from. Could anyone help me out here?

Thanks in Advance,

TFM

 

[Dick]

E=p^2/(2m) in operator notation since the potential in the box is 0. What's the operator p? Apply that to your wavefunction. (p^2/(2m))psi=E*psi.

 

[TFM]

Okay, so would the operator p be:

\int\Psi^*(\frac{\hbar}{i}\frac{\partial}{\partial x})\Psi dx

Or, since it s a operator, is it:

\hat{p} = -i\hbar \frac{\partial}{\partial x}

?

 

[TFM]

Okay, I have checked out the \hat{p} version, and I think it has given me the right answer:

E = \frac{\hat{p}^2}{2m}

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

E\psi(x) = \frac{\hat{p}^2}{2m} \psi(x)

E\psi(x) = \frac{-i^2 \hbar^2}{2m} \frac{\partial^2}{\partial x^2}Asin(\frac{n\pi}{L}x)

E\psi(x) = \frac{\hbar^2}{2m} A\frac{\partial^2}{\partial x^2}sin(\frac{n\pi}{L}x)

Do the first differential:E\psi(x) = \frac{\hbar^2}{2m} A\frac{n\pi}{L}\frac{\partial}{\partial x}cos(\frac{n\pi}{L}x)

And the second:

E\psi(x) = \frac{\hbar^2}{2m} A\frac{n^2\pi^2}{L^2}[-sin(\frac{n\pi}{L}x)]

Slightly rearrange:

E\psi(x) = -\frac{\hbar^2}{2m} \frac{n^2\pi^2}{L^2}[Asin(\frac{n\pi}{L}x)]

Since:

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

We can take out psi from both sides:

E = -\frac{\hbar^2}{2m} \frac{n^2\pi^2}{L^2}

rearrange very slightly:

E = -\frac{\hbar^2n^2\pi^2}{2mL^2}

And this is the same as the answer that is given in the book. Does this look okay?

Assuming it is okay, could you advise me what o do for the third part?

Use the results of (i) and (ii) and write down the complete time-dependent wave function Ψn(x,t) for the nth stationary state in this potential.

How do we get the time dependent equation from these two equations?

 

[Dick]

That looks ok, but you've got an extra minus sign hanging around. E should come out to be positive. Shouldn't the -i^2 at the beginning be +i^2? Now write down the time dependent Schrodinger equation. You've got that H|psi>=E|psi>. You just have to attach the correct time dependence.

 

[TFM]

Well, if:

\hat{p} = -i\hbar \frac{\partial}{\partial x}

p hat squared should give:

i^2 = -1

-*-1 = 1, so that does still give


E\psi(x) = \frac{\hbar^2}{2m} A\frac{\partial^2}{\partial x^2}sin(\frac{n\pi}{L}x)

but to get rid of the final minus, it should be


E\psi(x) = -\frac{\hbar^2}{2m} A\frac{\partial^2}{\partial x^2}sin(\frac{n\pi}{L}x)

I am not sure what has gone wrong there...?

So now I need to use the time dependent Schrodinger equation, I need to use:

i\hbar \frac{\partial \phi}{\partial x} = -\frac{\hbar}{2m}{\partial^2 \phi}{\partial x^2} + V\phi

Does this:

H|psi>=E|psi>

mean that the Energy eigenvector I just calculated is equal to the Hamiltonian operator?

 

[Dick]

(-i)^2=(-1)^2*i^2=1*(-1)=(-1). I'm not sure how you are making it a plus one. And, yes, p^2/(2m) is the Hamiltonian operator.

 

[TFM]

OKay thanks for that.

So, the Schrödinger Equation with the Hamilton Operator is:

\hat{H} = \frac{\hat{p}}{2m} + V

Gives Schrödinger's Equation to be:

i\hbar \frac{\partial \phi}{\partial x} = \hat{H}\phi

and now we have just said that:

E = -\frac{\hbar^2n^2\pi^2}{2mL^2} is the Hamilton operator. So:

i\hbar \frac{\partial \phi}{\partial x} = -\frac{\hbar^2n^2\pi^2}{2mL^2}\phi

and

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

Put together:

i\hbar \frac{\partial}{\partial x}Asin(\frac{n\pi}{L}x) = -\frac{\hbar^2n^2\pi^2}{2mL^2}Asin(\frac{n\pi}{L}x)

Okay so far?

 

[Dick]

Why do you have d/dx on the left side of the Schrodinger equation? Shouldn't it be d/dt? And you should keep in mind that the psi will have a time dependent factor as well. Write it as psi(x)*f(t). f(t) is what you are looking for. 'E' isn't the Hamiltonian operator, it's the 'eigenvalue' of the Hamiltonian (the energy). And you've still got that pesky (-1) in front of the energy. Why won't it go away? Finally your equation will come out to be similar to the ordinary differential equation df(t)/dt=k*f(t). What are the solutions to that?

 

[TFM]

okay, that was a typo should have been dt:

and with the f(t):

i\hbar \frac{\partial (\phi* f(t)}{\partial t} = \hat{H}(\phi*f(t)

and:

E = \frac{\hbar^2n^2\pi^2}{2mL^2}

This is the eigenvalue of the Hamiltonian Operator.

The Hamiltonian Operator is:

<br /> \hat{H} = \frac{\hat{p}}{2m} + V(x) <br />

So having a look, to get the Hamiltonian, I need:

\hat{H}\phi = E\phi

now we know that:

E = \frac{\hbar^2n^2\pi^2}{2mL^2}

And I assume phi is: <br /> \phi(x) = Asin(\frac{n\pi}{L}x) <br />

so:

\hat{H}Asin(\frac{n\pi}{L}x) = \frac{\hbar^2n^2\pi^2}{2mL^2}Asin(\frac{n\pi}{L}x)

Is this okay now?

Eidt: I assume not, since it is giving the same result for the Hamiltonian as before...

 

[Dick]

That's fine. But you've already done that.
<br /> i\hbar \frac{\partial (\phi(x) f(t))}{\partial t}= i\hbar \phi(x) \frac{\partial f(t)}{\partial t}= \hat{H}(\phi(x) f(t))=E \phi(x) f(t) <br />

You want to cancel out the phi(x) and solve for the f(t) part.

 

[TFM]

Okay so:

i\hbar \frac{\partial (\phi* f(t)}{\partial t} = \hat{H}(\phi*f(t)

goes to:

i\hbar\phi(x) \frac{\partial ( f(t)}{\partial t} = \hat{H}(\phi*f(t)

So now we need to replace the Hamiltonian with the Energy:

i\hbar\phi(x) \frac{\partial ( f(t)}{\partial t} = \frac{\hbar^2n^2\pi^2}{2mL^2}(\phi*f(t)

Is this right now?

 

[Dick]

Yes. It's right. Now you've got phi(x) on both sides. Cancel them out. You are just trying to find f(t) now. The 'time dependence'.

 

[TFM]

Excellent so:

i\hbar\phi(x) \frac{\partial f(t)}{\partial t} = \frac{\hbar^2n^2\pi^2}{2mL^2}\phi*f(t)

So cancel out the phi's:

i\hbar \frac{\partial f(t)}{\partial t} = \frac{\hbar^2n^2\pi^2}{2mL^2}f(t)

now rearrange so:

\frac{\partial f(t)}{\partial t} = \frac{\hbar^2n^2\pi^2}{i \hbar2mL^2}f(t)

and:

\frac{\partial f(t)}{\partial t} = \frac{\hbar n^2\pi^2}{i 2mL^2}f(t)

So would the solution be:

f(t) = e^{\frac{\hbar n^2\pi^2}{i 2mL^2}t}

Edit: that doesn't seem very cl;ear, it should be:
e^ft

where:

f = \frac{\hbar n^2\pi^2}{i 2mL^2}

?

 

[Dick]

I think so, yes. But it would probably be a lot more illuminating to write that as exp(-i*t*E_n/hbar) (where E_n is your energy), without trying to simplify it further.

 

[TFM]

Okay, so:

e^ft, where :

f = \frac{\hbar n^2\pi^2}{i 2mL^2}

this can be simplified by:

E = \frac{\hbar^2n^2\pi^2}{2mL^2}

to:

f = \frac{E_n}{\hbar}

thus:

f(t) = e^{(E_n/\hbar)t}

not sure where the -i is from?

 

[Dick]

<br /> f = \frac{E_n}{i \hbar}<br />
You just forgot the i. And 1/i=(-i).

 

[TFM]

Oh, I see now, thanks for pointing that out.

so f(t) = e^{-(iE/\hbar)t}

so is that the answer now for part (iii)?

 

[Dick]

Oh, I see now, thanks for pointing that out.

so f(t) = e^{-(iE/\hbar)t}

so is that the answer now for part (iii)?

f(t) is just the time part. The complete wavefunction is phi(x)*f(t).

 

[TFM]

Okay so the final answer will be:

\phi*f(t)

Which is:

Asin(\frac{n\pi}{L}x)*e^{-(iE/\hbar)t}

 

[Dick]

Yes, but be sure to make it clear that the E depends on the n as well.

 

[TFM]

Okay so it should be:

Asin(\frac{n\pi}{L}x)*e^{-(iE_n/\hbar)t}

Okay I pretty sure I can do most of the next question, can I just confirma few things?:

for the nth states, calculate: \left\langle \hat{x} \right\rangle , \left\langle \hat{x^2} \right\rangle , \left\langle \hat{p} \right\rangle , \left\langle \hat{p^2} \right\rangle , \Delta \hat{x}, \Delta \hat{p}

Now I know that, firstly:

\p{hat} = -i\hbar \frac{partial}{\partial x}

and therefore:

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

\left\langle \hat{p^2} \right\rangle = \phi^* \hat{p}^2 \phi

and just insert the p-hat into the equations.

I also know:

\Delta \hat{p} = \sqrt{\left\langle \hat{p^2} \right\rangle - \left\langle \hat{p}^2 \right\rangle}

I assume this follows on with the x-hat, but I am slightly unsure what the \hat{x} actually represents...?

 

[Dick]

The x-hat operator just multiplies a function by x. And to get the expectation value you also have to integrate over all x. E.g. <x>=integral (phi)* x (phi)dx.

 

[TFM]

Okay, so then:

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

and

\left\langle \hat{x^2} \right\rangle = \phi^*x^2\phi

Is this right then?

 

[Dick]

Right, but like I said don't forget to integrate over x.

 

[TFM]

Okay so:

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

Are we doing this just for:

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

or for the whole time dependence as well:

Asin(\frac{n\pi}{L}x)e^{-\frac{iE_n}{\hbar}t}

I think it is for the whole thing because the conjugate f(x)* requires a i which reverses value?

 

[Dick]

The time dependence will cancel, yes. But don't forget you need to normalize phi(x). It's properly normalized if you pick A so that the integral of (phi)* (phi) is 1.

 

[TFM]

But to get the conjugate,

\phi^*

Don't you need an i in the equation, because phi hasn't:

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

?

 

[Dick]

Your phi is real, yes. But that doesn't make it hard. What's the complex conjugate of a real number?

 

[TFM]

Won't it just be the same number, since its real?

 

[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.