# Infinite square well centered at the origin

• Kaguro

#### Kaguro

Moved from a technical forum
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?

## Answers and Replies

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?

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.

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.

Nugatory
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

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##.

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?

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?)

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##.

But I still don't know ψ\psi.

Can you write down an expression for ψ?

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.

##\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.

Last edited:
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?

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}##

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.

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.

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...

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?

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.

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).

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##.

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)?

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.

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

This is a good point to emphasize.

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?

Last edited:
vanhees71 and PeroK
Okay, apart from having two "odds" in the final equation.

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

Hmm... that's odd...

Sorry.

vanhees71
Thanks to everyone for help! I'll learn lots of QM!

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.

vanhees71
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, atyy and PeroK
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 :-).

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?

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?

vanhees71
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?