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Why is this subject's homework due before the lecture on it?

In the most recent email I think the rest of the finite square well (the odd solutions) as well as chapter 3 sections 1,2 are required reading - sections 1 and 2 are a “if we get to them” type of thing for Wednesday.

It seems like you can write the inner product of two vectors just like you do the inner product of two functions, < α | β > and < f | g >. Is this unintentionally sloppy? I don't think that you can take the inner product of a function with a vector, or is it the case that functions are usually represented by regular letters and vectors use greek letters? That would certainly make sense when we use ψ as a vector in Hilbert space.

This is not sloppy notation. Quantum mechanics lies in the subspace of Hilbert space in which functions are square-integrable and finite everywhere. This is also a vector space, meaning that Hilbert space is a vector space of functions. Functions are vectors in the mathematical sense in that they satisfy certain properties. For instance, the space of polynomials is a vector space. Your intuition is right; the inner product of two functions is the inner product of two vectors. In this case, we have infinite-dimensional vectors called functions. The inner product with arrow-like vectors is used to find the projection of one vector along another, and the same is true for our inner product of functions. Depending on the basis of our functions (vectors in Hilbert space), the inner product will be a sum for discrete bases <math><\alpha|\beta>=\sum_{i=1}^{N}a^{*}_{i}b_{i}</math> and an integral for continuous bases <math><\alpha|\beta>=\int_{A}^{B}a^{*}(x)b(x)dx</math>. This makes sense if you consider the projection of one function along another function which is orthogonal to first; the inner product will be zero, just as the dot product for perpendicular arrow vectors is zero. I hope this helps.

From today's discussion about the two square wells separated by some distance b. When the Energy of the systems is calculated as b approaches 0 and when b approaches infinity we get that the total energy is greater when the wells are far apart. Is the increase in the energy do to the fact that energy is brought into the system when the separated distance b is increased? I ask this because it makes more sense to me that the energy would increase as b goes to 0 not the opposite

Why does it make more sense that bringing the wells together would increase the energy? Are you thinking in terms of what might be creating the potential? Remember that we are only looking at the energy of a particle affected by this potential, not the underlying system creating it. The only way to understand the energy of the particle is to look at what kinds of wavefunctions Schrodinger's equation will admit for this potential. <math>E\propto\frac{1}{\lambda^2}</math>, so the shorter the wavelength, the higher the energy. When b=0, we have a finite square well of size 2a. When b»a, we have two finite square wells of size a that are almost independent of each other. The two wells of size a restrict the wavefunction to shorter wavelengths than one well of size 2a, so the b»a case has a higher energy.

I think you are associating increased energy with decreased separation because you're thinking about the repulsion between the nuclei, which we aren't considering here. Anyways, we're considering the energy of the electron in the bound state, not the energy of the molecule as a whole.

You can also think of this system as two electrons, in that for an electron to be in an excited state (further from 'zero') it has to have more energy – and for it to be farther away from another electron it also has to have more energy. I don't think this is completely correct, but as a memory tool, you could think of how when the electron comes back to ground it releases a photon, or energy. In the same way, when two wells come together, they can be at lower energy. When the second well comes closer to the first well it can “release” energy because this lower state requires less energy.

The double square well potential problem of today's discussion brought a question to my mind. If two square wells are separated by a some distance the wave functions of the bound energy states will be apparent in both wells. Lets say that there is another, empty, square well at an infinite distance away from these two wells, such that, initially, it has no effect on the wave functions present in the original two wells. Now let one of the square wells be moved away from the other square well over time, that is the distance between the wells is increasing, such that the moving well is approaching the empty square well. Eventually, the moving well will be close enough to the empty well that it will have a nonzero tunneling factor and its wave function will change accordingly to fit this new potential scenario. My question is will the wave function still be present in the stationary square well? And if it is, must the wave function portion present in that well change when the moving well approaches the empty well? And if it does, does this change occur instantaneously when the tunneling factor becomes nonzero?

I need a clarification on Problem 3.1(b) from the homework. Are the conditions the problem refers to only equations A.19, A.20, and A.21?

Yes. At least that's how I did the problem and it seemed to make sense.

A.(19-21) provide the only conditions required for an inner product. If these are met, you have satisfied the conditions for an inner product.

I have a quick question on the basic finite square well problem. In the book, k is defined in two different ways, depending on if E < 0 or E > 0 (Just looking at the points ouside of the well). From what I can tell, it looks like they are just setting up the Schroedinger Equation and rearranging constants. However, should be the same equation for both situations, which, solving, should give <math>k = \frac{\sqrt{-2mE}}{\hbar}</math>. I think I can pick out that they are saying the general solution for E < 0 has no i in the exponential (because it comes from k) and in the E > 0 situation there is an i (because they define k as not having a -ve inside the square root), but how do they get that solution of <math>k = \frac{\sqrt{+2mE}}{\hbar}</math>, since there is a -ve sign in the Schroedinger Equation?

As you already know, when V(x)=0, we have

<math>\frac{d^2\psi}{dx^2} = -\frac{\sqrt{2mE}}{\hbar}\psi</math>

For this explanation, consider the negative sign to be part of the coefficient of <math>\psi</math>.

When the energy is less than zero, that makes the coefficient of <math>\psi</math> greater than zero. In that case, no matter how you assign your constants, the value of the coefficient is positive. So you don't have a wave equation; your solutions to the differential equation don't need an i in order to give you positive a coefficient of <math>\psi</math>.

On the other hand, when the energy is greater than zero, then the coefficient of <math>\psi</math> is less than zero, and you do have a wave equation; you need an i in the solution to make the coefficient negative.

This reasoning can be extended to any region where V(x) is not equal to zero. You just have to look at whether [E-V(x)] is greater or less than zero.

If I goofed up my explanation, please somebody fix it. Thanks.

It's as simple as k is chosen to be positive and real by convention. For the E<0 case, the k you have is real and positive. For E>0, the second k you have listed is real and positive. If you are solving the Schroedinger Equation for an area with a nonzero potential, remember to include the V in your determination of whether k is positive or negative.

I have a notation question. The book gives a fourier trick used to find coefficients for functions that are orthonormal in Hilbert space. I'm a bit confused by the notation used, it gives Cn=<fn|f>. What would this look like in matrix form?

if you take a look at equation 3.10, then Cn=<fn|f> would make sense. think about Cn is a matrix of constant and fn is a collection of all solutions. <math>C = \begin{Bmatrix} c_1
c_2
.
.
c_n

\end{Bmatrix}  </math>  <math> fn = 

\begin{Bmatrix} f_1
f_2
.
.
f_n

\end{Bmatrix}  </math>

then corresponding each Cn with the element in the matrix, then use orthonormal rule to evaluate each integral

<math> \begin{Bmatrix} c_1
c_2
.
.
c_n

\end{Bmatrix} =<f_n|f> =\int f_nf\, dx =   \begin{Bmatrix}
 \int f_1*f\, dx          \\
 \int f_2*f\, dx          \\
  \int f_n*f\, dx=\delta_n_m\\
  .\\
  .\\
 \int f_n*f\, dx
 \end{Bmatrix}</math>

Also in response to Physics4dummies' question, with a question of my own: From what I understand, this problem is supposed to crudely represent an electron's state when positioned around two nuclei. Since the energy of the electron in the ground state goes up as the separation between the wells goes up, and since it wants to exist at lower energy, a force exists to push the two nuclei (wells) together. In the first excited state, the energy is lowered as the separation increases, so the force acts to pull the nuclei apart. My question is why does the energy of the electron go down as the wells are pulled apart in the second state, but up when it is in the first state?

It is a probably fundamental question.. When we got the general solution(<math>\Psi_I_I(x)=B*exp(ikx) + C*exp(-ikx))</math> in the region II for the finite square well which is from the lectures on last Mon & Wed, B = -C since <math/> Psi_II(0) = 0</math>. I don't understand why <math>\Psi_I_I(x=0)</math> becomes “zero”.

There are two different approaches about this problem. The first one is Odd case and the other is Even case. In the lecture, Professor considered of Odd case, which is asymmetric. The Shrodinger equation of the second region can be expressed like B`sin(kx). Therefore we will get B = -C since <math/> Psi_I_I(0) = 0</math>.

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