Infinite square well centered at the origin

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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?
 
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  • #27
PeroK
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Okay, apart from having two "odds" in the final equation.
 
  • #28
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Okay, apart from having two "odds" in the final equation.

Hmm... that's odd...

Sorry.:sorry:
 
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  • #29
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Thanks to everyone for help! I'll learn lots of QM!
 
  • #30
PeterDonis
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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.
 
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  • #31
George Jones
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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)##.
 
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  • #32
vanhees71
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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 :-).
 
  • #33
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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?
 
  • #34
PeroK
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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?
 
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  • #35
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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?
 
  • #36
PeroK
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To get solution for well from 0 to L ,I should make the transformation x →\to x+L/2
You must be careful with this technique. Which ##x## is which?

In fact, you ended up with them the wrong way round. Instead, try:

##x' = x + L/2##

Where ##x'## is the new variable that runs from ##0## to ##L##.

(Note that this means that ##x = x' - L/2##. Which is not what you did.)
 
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  • #37
vanhees71
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Also note that phase factors don't play any role. Defining the system to live in ##[0,L]## the boundary value at ##x=0## tells you ##\psi(0)=0##, which leads to the ansatz ##\psi(x)=A \sin(k x)## for the energy eigenfunctions, then the other boundary condition tells you
$$\psi(L)=A \sin(k L)=0,$$
from which you get ##k \in \pi \mathbb{N}/L##. Thus the complete set of eigenfunctions is
$$\psi_n(L)=A \sin \left(\frac{n \pi x}{L} \right).$$
The factors ##(-1)^j## are phase factors which you can simply neglect, because they have no physical significance.
 
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  • #38
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The factors (−1)j(-1)^j are phase factors which you can simply neglect, because they have no physical significance.
Yes, after understanding that x = x' - L/2 will be the correct transformation, I tried it again and got everything correct except this ##(-1)^{n/2}##. Now if you say that it carries no significance and can be neglected, I have arrived at the correct answer! Thank you!
 
  • #39
PeroK
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Yes, after understanding that x = x' - L/2 will be the correct transformation, I tried it again and got everything correct except this ##(-1)^{n/2}##. Now if you say that it carries no significance and can be neglected, I have arrived at the correct answer! Thank you!
To explain this a bit more.

What you got was a specific case of:

##\psi(x') = \alpha_n \sqrt{\frac{L}{2}} \sin (\frac{n\pi x'}{L})##

Where ##\alpha_n## is a complex number of unit modulus (sometimes called the phase factor). In this case, you had ##\alpha_n = \pm 1##, depending on ##n##.

This is perfectly valid for a set of eigenfunctions. But, you are free to choose the phase factor to be simply ##1## in each case. Hence, you have the simplest set of solutions:

##\psi(x') = \sqrt{\frac{L}{2}} \sin (\frac{n\pi x'}{L})##:

Note that can always do the opposite as well, and add any phase factors you want. For example, we could take your original solution:

##\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} ##
And, we could replace this with an alternative, equally valid set of eigenfunctions:

##\psi (x) =\begin{cases} i\sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ (e^{i\theta_n})\sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases} ##

This has all the same properties as your original solution. The only different is that we've added a phase factor to every function.

This is an important point, because when you have a set of eigenfunctions it can be very useful to choose a phase factor in a particular way. Often that some arbitrary complex number is real.

In any case, the point to remember is that you did not find the only possible unique set of solutions. You found a set of solutions, unique only up to the phase factor. You are free, therefore, to simplify or complicate your solution by changing the phase factors any way you like.
 
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