Infinite square well centered at the origin

[Kaguro]

The problem is:
Solve the time independent Schrodinger Equation for infinite square well centered at origin. Show that the energy is same as in the original case(well between x=0 and x=L). Also show that the solution to the this case can be obtained by setting x to x-L/2 in $\psi$ in the original one.

Now, in the original problem, E = $\frac{n^2\pi^2ħ^2}{2mL^2}$.

Here, I got my answer as 4 times that value:

From the Schrodinger Equation, put V=0 in (-L/2,L/2):
$\frac{d^2\psi}{dx^2}$ + $\frac{2mE}{ħ^2}\psi$ = 0

So,
$\psi$ = Acos(kx) + Bsin(kx) where $k^2 = \frac{2mE}{ħ^2}$

Now the boundary conditions require $\psi$ be 0 at x=-L/2 and x=+L/2
Putting these and solving them I get:
0 = Acos(kL/2) - Bsin(kL/2) and 0 = Acos(kL/2) + Bsin(kL/2)

From them:
tan(kL/2) = A/B and also A/B = -tan(kL/2)

which means: 2tan(kL/2) = 0
and so kL/2 = n$\pi$

From the definition of k, I have:
E = $\frac{2n^2\pi^2ħ^2}{mL^2}$

What mistake did I make?

 

[PeterDonis]

From them:
tan(kL/2) = A/B and also A/B = -tan(kL/2)

This assumes that B is nonzero. What if B is zero?

 

[Kaguro]

This assumes that B is nonzero. What if B is zero?

Actually I am having trouble finding the constants A and B. I tried various ways and got 4 possibilities:
A=0,B=0(not together), cos(kL/2) = 0, or sin(kL/2)=0.

Except the condition that A and B together must not be zero, and the condition that kL/2 can't have two different values, I don't see any condition which will tell me which constant to set to 0.

 

[PeroK]

Actually I am having trouble finding the constants A and B. I tried various ways and got 4 possibilities:
A=0,B=0(not together), cos(kL/2) = 0, or sin(kL/2)=0.

Except the condition that A and B together must not be zero, and the condition that kL/2 can't have two different values, I don't see any condition which will tell me which constant to set to 0.

Either $A = 0$ or $B = 0$. You need to solve both cases to get the full list.

You solved for $A = 0$ only, hence got every other solution, hence your $n$ was out of step with the full set of energy states.

 

[Kaguro]

Either $A = 0$ or $B = 0$. You need to solve both cases to get the full list.

You solved for $A = 0$ only, hence got every other solution, hence your $n$ was out of step with the full set of energy states.

If I do that, I get two different values of k:
Set A=0:
k = 2n$\pi$/L

Set B=0:
k = (2n-1)$\pi$/L

 

[PeroK]

If I do that, I get two different values of k:
Set A=0:
k = 2n$\pi$/L

Set B=0:
k = (2n-1)$\pi$/L

There's nothing wrong with that. That's an either/or. There would be nothing wrong with finding that, say: $k =0$ or $k =1$.

 

[Kaguro]

There's nothing wrong with that. That's an either/or. There would be nothing wrong with finding that, say: $k =0$ or $k =1$.

Well... then can you tell me how to proceed from here? I have 2 possible values of k and so using the definition of k in terms of E I can get two possible values of E:

E = $\frac{2n^2\pi^2ħ^2}{L^2}$
Or
E = $\frac{(2n-1)^2\pi^2ħ^2}{2mL^2}$

Which one to take? And why?

 

[PeterDonis]

I have 2 possible values of k

Both of which are actually special cases of one general formula for $k$. Can you see what it is?

(Hint: $2n$ is even and $2n - 1$ is odd. When you combine the set of even numbers with the set of odd numbers, what do you get?)

 

[Kaguro]

Both of which are actually special cases of one general formula for $k$. Can you see what it is?

(Hint: $2n$ is even and $2n - 1$ is odd. When you combine the set of even numbers with the set of odd numbers, what do you get?)

Okay okay... I can take just n. So this works up pretty well and I get the correct value for energy. But I still don't know $\psi$.

 

[Vanadium 50]

But I still don't know ψ\psi.

Can you write down an expression for ψ?

 

[PeroK]

Well... then can you tell me how to proceed from here? I have 2 possible values of k and so using the definition of k in terms of E I can get two possible values of E:

E = $\frac{2n^2\pi^2ħ^2}{L^2}$
Or
E = $\frac{(2n-1)^2\pi^2ħ^2}{2mL^2}$

Which one to take? And why?

I think you lost sight of what you were trying to do. You were looking for ALL solutions to the equation:

$\psi'' + k^2 \psi =0$

With the boundary conditions that $\psi(L/2) = \psi(-L/2) = 0$

First, you found a sequence of solutions for odd positive integers $n$:

$k = n \pi/L$

and

$\psi(x) = A \cos(\frac{n\pi x}{L})$

And, you found a sequence of solutions for even positive integers $n$:

$k = n \pi/L$

and

$\psi(x) = B \sin(\frac{n\pi x}{L})$

It's not clear why you think that you can only have solutions for odd $n$ or even $n$ but not both. If you try these functions in the origin equation, you'll see that both sets of functions are indeed solutions.

In any case, you clearly have a solution for every positive integer.

 

[Kaguro]

Can you write down an expression for ψ?

$\psi$= Acos (kx) + Bsin (kx).
I now know k but not A or B.

 

[PeroK]

$\psi$= Acos (kx) + Bsin (kx).
I now know k but not A or B.

In general, $A$ and $B$ can be any complex numbers. But, wavefunctions must be normalised, so you need to calculate the magnitude of $A$ and $B$ to get a well-defined eigenfunction in each case.

 

[PeterDonis]

I now know k but not A or B.

For any particular value of $k$ (i.e., any particular $n$), won't one of $A$ or $B$ be zero?

 

[Kaguro]

In general, $A$ and $B$ can be any complex numbers. But, wavefunctions must be normalised, so you need to calculate the magnitude of $A$ and $B$ to get a well-defined eigenfunction in each case.

If I try to normalize then I get this:
$A^2+B^2=\frac {2}{L}$

 

[Kaguro]

For any particular value of $k$ (i.e., any particular $n$), won't one of $A$ or $B$ be zero?

If you mean at the boundaries then I would say it depends on whether n is odd or even.

 

[PeterDonis]

If you mean at the boundaries then I would say it depends on whether n is odd or even.

The values of $A$ and $B$ for a particular value of $n$ can't be different at the boundaries than elsewhere; the solution for a particular value of $n$ has to have the same $A$ and $B$ everywhere within the well.

 

[Kaguro]

The values of $A$ and $B$ for a particular value of $n$ can't be different at the boundaries than elsewhere; the solution for a particular value of $n$ has to have the same $A$ and $B$ everywhere within the well.

Yes but I don't have any information within the well. Only that V=0...

 

[PeterDonis]

I don't have any information within the well.

So what? Did you read what I said? Did you understand that, since the boundary conditions are sufficient to find $A$ and $B$ at the boundaries for each $n$ (one of the two is zero for every $n$, so the other can be found easily by normalization), that gives you the solution for each $n$ within the entire well?

 

[Kaguro]

The boundary conditions are sufficient to find $A$ and $B$ at the boundaries for each $n$ (one of the two is zero for every $n$

Evidently I don't see how...

I think this is about time I give up and look up the solution from somewhere. Thanks for help.

 

[PeterDonis]

Evidently I don't see how...

You already knew that either $A = 0$ or $B = 0$ for every $n$ when you made your post #5. So for every $n$, you only have one nonzero term, and you can find its coefficient ($A$ or $B$, whichever is nonzero for that value of $n$) by normalization (all you have to do is take the formula you gave in post #15 and remove whichever of $A$ or $B$ is zero for that $n$). So for every $n$ you know the exact solution.

I think this is about time I give up

You should not give up now, when you are almost there (you've already done all the pieces, as above, you just need to put them together).

 

[PeroK]

If I try to normalize then I get this:
$A^2+B^2=\frac {2}{L}$

That looks right. But, remember that for every solution either $A = 0$ or $B = 0$.

 

[hutchphd]

That looks right. But, remember that for every solution either $A = 0$ or $B = 0$.

Does the OP understand that all of the A and B should really carry a subscript n (or k) and that specifying a particular solution for a given problem requires more information (perhaps the wavefunction everywhere at t=0)?

 

[PeterDonis]

specifying a particular solution for a given problem requires more information (perhaps the wavefunction everywhere at t=0)?

The problem asks for solutions to the time-independent Schrodinger Equation, so it's asking for energy eigenstates, not a general time-dependent state (which would be an arbitrary linear combination of energy eigenstates). Enough information is given to find the energy eigenstates.

 

[PeterDonis]

all of the A and B should really carry a subscript n

This is a good point to emphasize.

 

[Kaguro]

After giving the problem some rest when I looked at it again, I found that I was desperately trying to get a single function describe everything. Now that I've understood that it's not possible, I can easily present a piecewise defined function.

To summarise:
Write the TISE. Set V=0.
Find ψ= Acos (kx) + Bsin (kx)
Take help from boundary conditions to get:
0 = Acos(kL/2) + Bsin(kL/2) and 0= Acos(kL/2) - Bsin(kL/2)

This means Either A=0 or B=0.
Set A=0:
k = 2nπ/L

Set B=0:
k = (2n-1)π/L

So in general k = nπ/L
get energy from there: $E_n= \frac{n^2\pi^2ħ^2}{2mL^2}$

Normalise to get $A^2+B^2=\frac{2}{L}$
Then write the boundary condition again:
0 = $Acos(n\pi/2) + Bsin(n\pi/2)$

If n is odd:
0 = B*1 or B=0
Then $\psi = \sqrt{ \frac{2}{L}}cos(n\pi x/L)$

Similarly,
If n is even:
Then $\psi = \sqrt{ \frac{2}{L}}sin(n\pi x/L)$

So:
$\psi (x) =\begin{cases} \sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ \sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases}$

Is all good?

 

[PeroK]

Okay, apart from having two "odds" in the final equation.

 

[Kaguro]

Okay, apart from having two "odds" in the final equation.

Hmm... that's odd...

Sorry.

 

[Kaguro]

Thanks to everyone for help! I'll learn lots of QM!

 

[PeterDonis]

I found that I was desperately trying to get a single function describe everything. Now that I've understood that it's not possible, I can easily present a piecewise defined function.

Just to clarify terminology: you were never looking for a single function, because you have an infinite number of possible values of $n$, and each value of $n$ leads to a different function. Each of these functions describes a possible energy eigenstate for the system.

What you were looking for was a single general form that all of these functions would take--i.e., a single expression with $n$ appearing in it, that would describe all of the possible energy eigenstates. What you have given now is not a "piecewise defined" function (that would be a single function whose form was different for different ranges of $x$). It's two expressions, one for odd $n$ and one for even $n$, that together describe all of the possible energy eigenstates. In other words, your two expressions describe an infinite number of possible functions of $x$, one for each $n$, and each function (i.e., each specific value of $n$) is a possible energy eigenstate for the system.

 

[George Jones]

I probably shoudn't do this, because it might confuse the issue ... but, here is another way to do this problem assuming the solution for a well from $x=0$ to $x=L$ is known. Use translation of functions, i.e., stuff like $g\left(x\right) = f\left(x-a\right)$ and $h\left(x\right) = f\left(x+a\right)$.

 

[vanhees71]

That's great. Then you can also show that this of course doesn't change anything concerning the physics, since it's just a different choice of where you set the 0 point of your (spatial) reference frame. This "change of description" is formally given by a corresponding unitary transformation (a translation, which is a fundamental symmetry of Galilean spacetime), and thus indeed nothing is changed physicswise, just our description has changed :-).

 

[Kaguro]

I probably shoudn't do this, because it might confuse the issue ... but, here is another way to do this problem assuming the solution for a well from $x=0$ to $x=L$ is known. Use translation of functions, i.e., stuff like $g\left(x\right) = f\left(x-a\right)$ and $h\left(x\right) = f\left(x+a\right)$.

This sounds interesting! Can you please elaborate how to get the solution of centered well from the normal one?

Exactly where do I substitute these new limits?

 

[PeroK]

This sounds interesting! Can you please elaborate how to get the solution of centered well from the normal one?

Exactly where do I substitute these new limits?

Sometimes the square well is taken to be from $0$ to $L$ and in this case the solution comes out slightly more easily than it does in the cases of $-L/2$ to $L/2$.

Since you now have the solution for a well from $-L/2$ to $L/2$, you can get the solutions for a well from $0$ to $L$ using a change of coordinates.

Can you see how to do that?

 

[Kaguro]

Sometimes the square well is taken to be from $0$ to $L$ and in this case the solution comes out slightly more easily than it does in the cases of $-L/2$ to $L/2$.

Since you now have the solution for a well from $-L/2$ to $L/2$, you can get the solutions for a well from $0$ to $L$ using a change of coordinates.

Can you see how to do that?

I have $\psi= \sqrt{\frac{2}{L}} cos(n\pi x/L)$ if n is odd and
$\psi= \sqrt{\frac{2}{L}} sin(n\pi x/L)$ if n is even

To get solution for well from 0 to L ,I should make the transformation x $\to$ x+L/2

There for odd n:
$\psi= \sqrt{\frac{2}{L}} cos(n\pi (x+L/2)/L)$
$= \sqrt{\frac{2}{L}} cos(n\pi x/L)cos(n\pi /2) - sin(n\pi x/L)sin(n\pi /2)$
$=-\sqrt{\frac{2}{L}}sin(n\pi x/L)$

For even n:
$\psi= \sqrt{\frac{2}{L}} sin(n\pi (x+L/2)/L)$
$= \sqrt{\frac{2}{L}} (sin(n\pi x/L)cos(n\pi /2) + cos(n\pi x/L)sin(n\pi /2) )$
$= \sqrt{\frac{2}{L}}(-1)^{n/2} sin(n\pi x/L)$

How to combine them?