# Finding the expected value of position in a Potential Well (1 Viewer)

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#### ja07019

1. The problem statement, all variables and given/known data
Hello today I am solving a problem where an electron is trapped in a potential well. I have a solved Schrodinger's Equation. I am having problems in figuring out what the wave function should be. When I solved the equation I got a complex exponential. I know I cannot use the complex exponential because when multiplied by its complex conjugate, the answer is simply 1. The book says that the real component is the symmetric solution while the imaginary component is the asymmetric solution. So I tried to calculate the expected value but got an answer that was different from the one given by the textbook. Note that I tried both the symmetric solution and the asymmetric solution. I am asked to calculate the average value of (z-<z>)2. The potential well is in only 1 dimension and it starts at -L and ends at L.

2. Relevant equations
<A> = ∫Ψ*AΨdx
Ψ=Aexp(ikz)
ψ(symmetric)=Acos(kz) and thus ψ* = Acos(kz)
ψ(asymetric) = Asin(kz) and ψ*Asin(kz)

I know that <z> = 0

3. The attempt at a solution
Here I tried to compute both of the following integrals,
<(z-<z>)2> = <z2> = ∫[Azcos(kz)]2 dz
and likewise I also tried to compute
<z2> = ∫[Azcos(kz)]2 dz.
Both of these are definite integrals. I assumed that they were integrated from -L to L. The answer given in the book is [1-(-1)n(6/(nπ)2)]*a23.
When I attempted the symmetric solution, (the one with the cosine term), I yielded a solution of L3[(2(nπ)2 +3)/(6(nπ)2)].
The asymmetric solution, which uses sine, is somewhat similar to this solution.

#### BvU

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Hello ja,

Ψ=Aexp(ikz)
$$\Psi = e^{ikx}$$ Satisfies the Schroedinger equation, but not the boundary conditions. It also isn't the full solution. (SE is 2nd order, so we expect to see 2 integration constants). Please show your work in a post

#### ja07019

Hello ja,

$$\Psi = e^{ikx}$$ Satisfies the Schroedinger equation, but not the boundary conditions. It also isn't the full solution. (SE is 2nd order, so we expect to see 2 integration constants). Please show your work in a post
Hello thanks or the reply. I should have been more specific. Yes I omitted the time component. But it is also a complex exponential. The book I am using solves via separation of variables leading to $\Psi = r(t)\psi$, where $\psi is a function of space. The book explicitly uses the two symbols and they define the expected value in terms of the position component of the wave equation. I do believe that I only have to take the expectation with respect to only the position probability. Then again, Hello ja, $$\Psi = e^{ikx}$$ Satisfies the Schroedinger equation, but not the boundary conditions. It also isn't the full solution. (SE is 2nd order, so we expect to see 2 integration constants). Please show your work in a post Hello thanks or the reply. I should have been more specific. Yes I omitted the time component. But it is also a complex exponential. The book I am using solves via separation of variables leading to$\Psi = r(t)\psi $, where$\psi is a function of space. The book explicitly uses the two symbols and they define the expected value in terms of the position component of the wave equation.

I do believe that I only have to take the expectation with respect to only the position wave function. Another point to make, the book apparently says that cosine is the symmetric solution while sine is the asymmetric solution of the position wave function. I am inclined to believe that this has to do with the imaginary components.

#### BvU

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Is the potential a well or a box ?
Please show your work in a post

#### ja07019

Is the potential a well or a box ?
It's a potential well.
Okay so first I solved SE for a potential of 0(the well).

Start by assuming that,
$$\Psi = r(t) \psi (z)$$
which leads to,
$$\Psi _t = r'(t) \psi (z)$$
and
$$\Psi _{xx} = r(t) \psi ''(z)$$
Now going back to Schrodinger's equation, in a single dimension,

$$\frac{-\hbar ^2}{2m} \frac{\partial ^2\Psi}{\partial z^2}= i\hbar \frac{\partial \Psi}{\partial t}$$
and the equation becomes
$$\frac{-\hbar ^2}{2m} r(t) \psi ''(z) = i\hbar r'(t) \psi(z)$$
which leads to an equation with separated variables
$$\frac{-\hbar }{i2m} \frac{r'(t)}{r(t)} = \frac{\psi '' (z)}{\psi (z)}= constant$$
This yields two ODES,
$$\frac{-\hbar }{i2m} r'(t) +Ar(t) = 0$$
which should be solved as a linear first order and yields the solution(assume the constant is related to energy)
$$r(t) = exp(-i\frac{E}{\hbar}t)$$
and
$$\psi ''(z) + A \psi(z) = 0$$
This is just a second-order ODE, a general solution involving the complex exponential due to non-roots. The solution here is,
$$\psi (z) = Aexp(ikz)+Bexp(-ikz)$$
These two terms obviously are conjugates, which makes sense as there could be a transmitted wave and a reflected wave.
So in a potential well(where V=0),
$$\Psi = exp(-i \frac{E}{\hbar}t)[Aexp(ikz)+Bexp(-ikz)]$$

Then I solved for a nonzero potential(the walls):
I once again separated the variables, only this time I applied a change of coordinates to eliminate the V. Essentially all is the same. Let
$$\Psi (z,t) = W(z,t)exp(Vt)$$
which leads to
$$\Psi _t = W_t exp(Vt) + VWexp(Vt)$$
and
$$\Psi _{xx} = W_{xx} exp(Vt)$$
This changes the current form of SE from:
$$\frac{- \hbar ^2}{2m} \Psi _{xx} + V \Psi = i \hbar \Psi _t$$
into:
$$\frac{- \hbar ^2}{2m} W_{xx} exp(Vt) + V Wexp(Vt) = i \hbar W_t exp(Vt) + VWexp(Vt)$$
After simplifying this simply becomes,
$$\frac{- \hbar ^2}{2m} W_{xx} + V W = i \hbar W_t + VW$$
and thus
$$\frac{- \hbar ^2}{2m} W_{xx} = i \hbar W_t$$
Which leads to a solution of
$$\Psi = exp[(-i \frac{E}{\hbar}+V)t]$$
which essentially shifts the wave function by a certain amount to accommodate the the potential.
This leads into ensuring continuity of the potential well so that the wave equation is continuous both in the in its current form and in the derivative form.
But here's the thing, the book throws away the reflected portion of the wave equation(seems obvious to me), and then breaks up the transmitted part into a symmetric and asymmetric solution, the symmetric being Acos(kz) and the asymmetric Asin(kz). I suppose it has to do with the complex part.
So back to the my question. If I take the whole equation, this simply leads to the product of the complex function and complex conjugate function reducing to 1.
Just to be clear, I am solving for <(z-<z>)2>. I know that <z> = 0 and thus I make the assumption that <(z-<z>)2> simply reduces into <z2[/SUB]>. It seems reasonable to me because I'd have to compute either way and it will yield 0 in all cases and it would eliminate everything, and with the definite integral at play, it would completely reduce any constants as well. Of course there is an issue of whether I could be wrong and this assumption does not hold.

#### BvU

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Homework Helper
$$\Psi = exp[(-i \frac{E}{\hbar}+V)t]$$ blows up with time. Can't be right.  I'm wrong . But the solution isn't correct

You have a 2nd order DE in $x$ so there should be 2 integration constants -- which you have: $A$ and $B$ within the well -- but I see no solution and no integration constants outside the well (in the region of the wall) and no continuity requirements ?

where an electron is trapped in a potential well
meaning $V < E < 0$ .

#### ja07019

$$\Psi = exp[(-i \frac{E}{\hbar}+V)t]$$ blows up with time. Can't be right.  I'm wrong . But the solution isn't correct

You have a 2nd order DE in $x$ so there should be 2 integration constants -- which you have: $A$ and $B$ within the well -- but I see no solution and no integration constants outside the well (in the region of the wall) and no continuity requirements ?

meaning $V < E < 0$ .
I didn't explicitly write it out, but continuity basically says that at z=-L, L, the two wave functions must be equal to each other. In addition to that

$$\frac{ \partial \Psi _{V=0} }{\partial z} = \frac{ \partial \Psi _{V=V_0} }{\partial z}$$

when z=-L,L. So from there I just have to find the derivative of each function, set them up equal to each other, and grab whatever values pop out. I do the same with the two functions, set them equal to each other at z=-L, L and then I have a solution.
The textbook deals with these values by using the Wronskian to find two solutions that are linearly dependent which brings about a certain condition for what the values that may be allowed for k.
The solution that I found matches the one in the textbook.
 I just realized there limits are -L/2 to L/2 for the length of the well.
But I seem to not be able to yield the proper solution.
Thanks for sticking around by the way!

#### BvU

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Homework Helper
What solutions do you have for $|z| \ge {L\over 2}$ ?

#### ja07019

What solutions do you have for $|z| \ge {L\over 2}$ ?
Alright so I'll begin by saying that I want the symmetric solution. Once again I'm going to iterate that I don't necessarily see why one solution is symmetric but I think it might have to do with the i.
Just to reiterate, I am evaluating
$$\langle (z- \langle z \rangle )^2 \rangle$$
I have already know that,
$$\langle z \rangle = 0$$
With this in mind, I simplify the statement above into
$$\langle z^2 \rangle$$.
To find the expected value, I know I have to normalize the limits, but I leave that till the end. I assume I am searching for the symmetric solution of the form,
$$\psi = Acos(kz)$$
and thus
$$\psi ^* = Acos(kz)$$
Thus, by the definition of the expectation of a random variable(z), I have
$$\int [Acos(kz)]z^2[Acos(kz)]dz = A^2 \int (zcos(kz))^2dz$$
I compute this integral by parts:
$$u=z^2$$
$$du = 2zdz$$
$$dv = cos^2(kz)dz = \frac{1+cos(2kz)}{2}dz$$
$$v = \frac{z}{2}+\frac{sin(2kz)}{4k}$$
$$\int (zcos(kz))^2dz = (z^2)[\frac{z}{2}+\frac{sin(2kz)}{4k}]- \int [\frac{z}{2}+\frac{sin(2kz)}{4k}]2zdz$$
$$=\frac{z^3}{2}+\frac{z^2sin(2kz)}{4k} - \int z^2dz - \int \frac{zsin(2kz)}{2k}dz$$
which leads to
$$=\frac{z^3}{2}+\frac{z^2sin(2kz)}{4k} - \frac{z^3}{3} - \frac{1}{2k} \int zsin(2kz)dz$$
I do yet another substitution,
$$q = z$$
$$dq = dz$$
and
$$dw = sin(2kz)dz$$
$$w = \frac{-cos(2kz)}{2k}$$
and thus
$$\int (zcos(kz))^2dz = \frac{z^3}{6}+\frac{z^2sin(2kz)}{4k} - \frac{1}{2k}[\frac{-zcos(2kz)}{2k} -\int \frac{-cos(2kz)}{2k}dz]$$
and this yields.
$$\int (zcos(kz))^2dz = \frac{z^3}{6}+\frac{z^2sin(2kz)}{4k} + \frac{zcos(2kz)}{4k^2} - \frac{sin(2kz)}{8k^3}$$
Finally applying the limits
$$(\frac{(L/2)^3}{6}-\frac{(-L/2)^3}{6})+(\frac{z^2sin(2(n \pi /L)(L/2))}{4k}-\frac{z^2sin(2(n \pi /L)(-L/2))}{4k}) + (\frac{zcos(2(n \pi / L)(L/2))}{4k^2} - \frac{zcos(2(n \pi /L)(-L/2))}{4k^2}) - (\frac{sin(2(n \pi /L)(L/2))}{8k^3}-\frac{sin(2(n \pi /L)(-L/2))}{8k^3})$$
which results in
$$\frac{L^3}{24}+(0-0)+\frac{1}{8k^2}(Lcos(n\pi)--Lcos(n\pi))-(0-0)$$
$$=\frac{L^3}{24}+\frac{\pm L}{4k^2}$$
However I shall now apply normalization to transform the function into a proper PDF:
$$\int \psi ^* \psi dz = 1$$
Here I will apply the normalization technique from -L/2 to L/2
$$\int A^2 cos^2(kz) dz = A^2 \int \frac{1+cos(2kz)}{2}dz = \frac{A^2}{2} \Big ( \int dz + \int cos(2kz)dz \Big ) = \frac{A^2}{2} \Big (z+ \frac{sin2kz}{2k} \Big ) = 1$$
The value of k along with L/2 will eliminate the sine portion leaving,
$$\frac{A^2}{2}\Big ( \frac{L}{2}-\frac{-L}{2} \Big ) = \frac{A^2}{4} (L+L) = \frac{LA^2}{2} = 1$$
Thus
$$A^2 = \frac{2}{L}$$

With normalization, the full solution is
$$\langle z^2 \rangle = \frac{L^2}{12}+\frac{\pm L}{2L(\frac{n \pi}{L})^2} = L^2\Big ( \frac{1}{12}+\frac{\pm 1}{2(n \pi)^2} \Big )$$
The solution given in the book is
$$\frac{a^2}{3}\Big [ 1 - (-1)^n \frac{6}{n^2 \pi ^2} \Big ]$$
I guess my question is now just about constants. Turns out I was using the wrong limits of integration.

#### BvU

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Homework Helper
Alright so
Your start suggests you are going to answer my question, but then you digress in a huge story. What is your complete problem statement ?
today I am solving a problem where an electron is trapped in a potential well
Now it is Wednesday and I haven't seen a decent solution to the TISE.
When I solved the equation I got a complex exponential
Are you referring to $$\Psi = \exp(-i \frac{E}{\hbar}t)[A\exp(ikz)+B\exp(-ikz)] \ ?$$because that isn't a complete solution , and as I tried to indicate, $\ \Psi = \exp[(-i \frac{E}{\hbar}+V)t]\$ is utterly wrong.
Please try not to jump to conclusions.

I know I cannot use the complex exponential because when multiplied by its complex conjugate, the answer is simply 1
That's the whole thing about the TISE: when the DE is separable, you get $H(\psi(z)) = E\psi(z)$ and $\Psi(z,t) = \psi(z)e^{-i\omega t}$ . With specific conditions for $\psi(r)$ which we need to explore in this exercise

The book says that the real component is the symmetric solution while the imaginary component is the asymmetric solution. So I tried to calculate the expected value but got an answer that was different from the one given by the textbook
I'm a little lost here. Can I see what precisely you are referring to ?

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#### BvU

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Homework Helper
Just to reiterate, I am evaluating
... I simplify the statement above into $⟨z^2⟩$
I sense a slight irritation, perhaps because I didn't go into that. I agree, but you need a workable solution first.

I don't want to aggravate things, but:
I assumed that they were integrated from -L to L.
Why ?

#### BvU

Science Advisor
Homework Helper
The book says that the real component is the symmetric solution while the imaginary component is the asymmetric solution
Ah, wait: You have $$Ae^{ikz} + B e^{-ikz} = \bar A \sin (kz)+\bar B \cos(kz)$$ with $\ \bar A = i(A - B) \$ and $\ \bar B = A + B\$ so if $A$ and $B$ are chosen as real (which is allowed -- do you know why ?) then the book statement is correct.

But the choices for $A$ and $B$ are not free, which leads me back to asking for the general solutions satisfying both $H(\psi(z)) = E\psi(z)$ and the continuity equations. They lead to equations that link $E$ and $L, V$. You don't have to solve those, but out come discrete possible $E_n$.
And the lowest $E_n$ (which we can call $E_0$) is for a symmetric $\psi(z)$.

Are you aware of the reason we separate into symmetric and antisymmetric solutions ?

#### ja07019

Alright so the full solution will obviously contain two parts, one where the voltage is zero and the other with an applied voltage
$$\Psi (z,t) = exp[(-i \frac{E}{\hbar}+V)t](Aexp(ikz)+Bexp(-ikz)]$$
and
$$\Psi (z,t) = exp[(-i \frac{E}{\hbar})t](Aexp(ikz)+Bexp(-ikz))$$
This is the answer the book has so I don't question much about it. I too arrived at this solution.

I am working inside a potential well now, so I take equation 1, I input the bounds of the potential well, and I equate that two equation 2 to ensure continuity. I find the derivative of both and do the same. From here I get the relation that
$$k = \frac{n \pi}{L}$$
Continuity is essentially that
$$\Psi_{V=V_0} (\frac{L}{2}, t) =\Psi_{V=V_0} (\frac{L}{2}, t)$$
$$\Psi_{V=V_0} (\frac{-L}{2}, t) =\Psi_{V=V_0} (\frac{-L}{2}, t)$$
and, the derivatives must satisfy that as well,
$$\Psi_{z,V=V_0} (\frac{L}{2}, t) =\Psi_{z,V=V_0} (\frac{L}{2}, t)$$
$$\Psi_{z,V=V_0} (\frac{-L}{2}, t) =\Psi_{z,V=V_0} (\frac{-L}{2}, t)$$
I know this is an essential step, but I just omitted these steps since I figured they only contribute to what k is.

Furthermore I make a little adjustment. The wave function above is the function for two waves in reality and can be interpreted physically. The second wave, $Bexp(-ikz)$ can be ignored. Note that this wave is just the first wave traveling in the reverse direction as indicated by the negative sign.
So the overall solution is
$$\Psi (z,t) = Aexp(-i \frac{E}{\hbar}t)exp(ikz)$$
The book then states that
$$\psi = Aexp(ikz)$$
which is just the space component. Notice that they use the uppercase, $\Psi$ for the full solution, and lowercase, $\psi$ for the space component.
The formula they give for the expectation is
$$\langle A \rangle = \frac{\int_{Vol} \psi ^* A \psi d(volume)}{\int_{Vol} | \psi |^2 d(volume)}$$
Okay so the book, then writes
$$Acos(kz)+iAsin(kz)$$
The book calls the first part, $Acos(kz)$, the symmetric solution and $Asin(kz)$ the asymmetric solution. I don't know why it is they do that, they don't really elaborate on that but I assume it has to do with the i?
In the recent post I just tried to evaluate $\langle (z- \langle z \rangle )^2 \rangle$

Now when you say I don't have the solution to TISE, do you mean that I lack the Boundary conditions and initial value? I didn't bother finding them because the book doesn't solve them. And I believe that matter is resolved when continuity and normalization are applied. I'm guessing continuity is a boundary condition and appears as something like this:
$$\Psi - \Psi _z = 0$$
Not exactly that statement, but something that comes from continuity required at the 'walls'. I have never solved SE before so I assumed no boundaries and it got me to the same point in the book.
Anyhow no irritation here I am glad someone even bothers to take a look. Thank you once again. Sorry if I sounded rude.

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#### ja07019

Ah, wait: You have $$Ae^{ikz} + B e^{-ikz} = \bar A \sin (kz)+\bar B \cos(kz)$$ with $\ \bar A = i(A - B) \$ and $\ \bar B = A + B\$ so if $A$ and $B$ are chosen as real (which is allowed -- do you know why ?) then the book statement is correct.

But the choices for $A$ and $B$ are not free, which leads me back to asking for the general solutions satisfying both $H(\psi(z)) = E\psi(z)$ and the continuity equations. They lead to equations that link $E$ and $L, V$. You don't have to solve those, but out come discrete possible $E_n$.
And the lowest $E_n$ (which we can call $E_0$) is for a symmetric $\psi(z)$.

Are you aware of the reason we separate into symmetric and antisymmetric solutions ?
Hmm I thought the second wave with B coefficient was removed because that would be a reflected wave from an incident wave.

No I do not know why one is symmetric and the other asymmetric.

#### ja07019

I quick search on StackExchange showed the solution for a square well as,
$$\sqrt{\frac{2}{a}}sin \frac{n \pi x}{a}e^{-i E_n t}$$
which the book describes as being the asymmetric solution. The entire time, the book worked on the symmetric solution so now it seems imperative to know what makes it be symmetric and asymmetric.

#### BvU

Science Advisor
Homework Helper
I thought the second wave with B coefficient was removed because that would be a reflected wave from an incident wave
Doesn't apply: that's for E>0 which doesn't give
an electron is trapped in a potential well
Re Stackexchange:
That would be for an infinitely deep well I suspect.

Re odd/even:
The potential is symmetric around $z = 0$, so $\psi^2$ should be symmetric too.
'Alternative' is using parity $P$: $\quad$ With $P(\psi(x)) = \psi(-x)$ and $P^2 = \mathcal I$

#### ja07019

Doesn't apply: that's for E>0 which doesn't give

Re Stackexchange:
That would be for an infinitely deep well I suspect.

Re odd/even:
The potential is symmetric around $z = 0$, so $\psi^2$ should be symmetric too.
'Alternative' is using parity $P$: $\quad$ With $P(\psi(x)) = \psi(-x)$ and $P^2 = \mathcal I$
I don't know anything about this information. Ah I see. This example is from a book on Electrical properties of Materials and this is about quantum tunneling in transistors. So I don't know anything about parity. But I do see how it appears symmetric. Cosine is indeed symmetric from -π to π. But I have a question, why is symmetry required here? Does it necessarily have to be that at z=0 a maximum likelihood is observed? What might force symmetry?
Thanks for the very helpful reply. I definitely didn't think about it that way. It really helps to visualize the problem.

#### BvU

Science Advisor
Homework Helper
why is symmetry required here
Not required. The Hamiltonian is symmetric, so when you flip (that is what the parity operator does) a solution, the result is also a solution.
However: the wave function itself is meaningless: only $\psi^2$ has a physical meaning. So solutions that are antisymmetric with $\psi(-x) = -\psi(x)$ are also good solutions (try it).

Does it necessarily have to be that at z=0 a maximum likelihood is observed?
So: no. Either a maximum (for a symmetric solution) or a minimum (for an antisymmetric solution).

All caused by the fact that the hamiltonian is second order in $dz$.

his example is from a book on Electrical properties of Materials and this is about quantum tunneling in transistors
Makes sense. But: we started working on a well. Let's finish that first, then look at the potential barrier.

#### ja07019

Not required. The Hamiltonian is symmetric, so when you flip (that is what the parity operator does) a solution, the result is also a solution.
However: the wave function itself is meaningless: only $\psi^2$ has a physical meaning. So solutions that are antisymmetric with $\psi(-x) = -\psi(x)$ are also good solutions (try it).

So: no. Either a maximum (for a symmetric solution) or a minimum (for an antisymmetric solution).

All caused by the fact that the hamiltonian is second order in $dz$.

Makes sense. But: we started working on a well. Let's finish that first, then look at the potential barrier.
Okay so does the solution for the wave correct? I have verified it with the book and it seems to be the solution that they too have. So now I'm at the part where I am trying to evaluate <(z-<z>)2> and this requires the wavefunction.

#### BvU

Science Advisor
Homework Helper
To avoid miscommuication: could you post the solution you are referring to ? I think I haven't seen it yet, but I can be mistaken.

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