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I know we kind of touched on this class, but I was confused about where the <math>\sqrt{1/2}</math> came from in the expression for a+ and a-?

What I understand is that from eq. 2.46 <math>a_-</math> and <math>a_+</math> are just the operators of the hamiltonian <math>H=\frac{1}{2m} [p^2+(m\omega x)^2]</math>. So the <math>\frac{1} {\sqrt{2}}</math> arise when you take the square root of the Hamiltonian. At least that is what I'm getting.

Yuichi

There is nothing magical about that <math>\sqrt{1/2}</math>. We can live without it. If we drop this factor in the definition of these operators, the relation between the Hamiltonian and these operators becomes <math>\hat{H}=\hbar\omega(a_+a_-/2+1/2)=\frac{\hbar\omega}{2}(a_+a_-+1)</math>, for example. Many relations involving these operators will be modified slightly, but the fundamental will be unchanged. This means that <math>a_+\psi_n\prop\psi_{n+1}</math>. But the normalization factor changes. I suggest that you will work these out to see what change and what do not.

I am having a hard time of understanding the different expressions for the total energy E. like for Bohr Model E is proportional to <math>\frac{1}{n^2}</math>, say <math>E_n=-E_0\frac{1}{n^2}</math>, for SHO <math>E_n=\frac{1}{2}hw(n+1)</math>, for infinite square well <math>E_n=E_0n^2</math>, anyone have a better explanation to get comfortable with them.

The different expressions for energy E are simply consequences of solving the time-independent Schrodinger's equation (2.5) for different assumed forms of the potential V (Coulomb's law - for Bohr's model, infinite well, SHO…). For a each of these forms of V, there is an infinite amount of solutions to equation 2.5, each corresponding to a function <math>\Psi (x)</math> and a value of E. On pages 26-27 you can find an explanation for why the variable E in equation 2.5 is the energy. The energy <math>E_n</math> is the energy corresponding to the solution numbered n.

In discussion 2 solutions, shouldn't the coefficients in the time-dependent solutions, to the example problem, be 1/sqrt(2)? I am confused as to why they are 1/sqrt(a).

Yuichi I almost thought that we made a mistake in the solution. Well, it was not the case. The normalized stationary wave functions is <math>\psi_n(x)=\sqrt{2/a}\sin n\pi x/a</math>. When you combine <math>A=\sqrt{1/2}</math> with the normalization factor in the stationary state wave function, <math>\sqrt{2/a}</math>, you get <math>\sqrt{1/a}</math>.

In discussion today, a lot of us had trouble solving the problem, which is also part of our homework. The main problem seemed to be our inability to perform a Fourier expansion of an arbitrary function. Ryo recommended looking at a calculus textbook, as well as the textbook for this class for an explanation, but since this seems to be such a widespread issue, perhaps it would be appropriate to cover Fourier expansions in more detail in class? Alternatively, someone who is comfortable with them could write a tutorial for the rest of us here.

Near the beginning of Section 2.3.2 of the book (“Analytic Method”) I am not following the jump between equation 2.76 and 2.77. How does one “'peel off'” the <math>h(\xi)</math> factor?

About HW2 problem 2.2, I do not think Yuichi gives a strict proof in Physics4101Hw2.pdf, which seems to me as an interpretation rather than a proof. I would like to show my idea why the function in that problem cannot be normalized. As <math>\psi</math> always has the same sign with its second derivative, first we can conclude that <math>\psi</math> must be real, because a complex number do not have 'sign'. Define <math>G(x)=|\psi(x)|^2=\psi(x)^2</math>, then <math>G(x)</math> is always <math>\ge 0</math>. <math>\frac{d^2G(x)}{dx^2}=2\psi(x)\frac{d^2\psi(x)}{dx^2}+2(\frac{d\psi(x)}{dx})^2</math>, obviously the second term is always <math>\ge0</math>, and the first term is also <math>\ge0</math> because <math>\psi(x)</math> and <math>\frac{d^2\psi(x)}{dx^2}</math> have same signs. Then <math>\frac{d^2G(x)}{dx^2}</math> is always <math>\ge 0</math>. This means in any <math>G(x)</math> intervals <math>[x_1, x_2]</math>, and any point <math>x_0\in[x_1, x_2]</math>, they must satisfy <math>G(x_1)+G(x_2)\ge2G(x_0)</math>. As <math>G(x)</math> is always <math>\ge 0</math>, there must exist a certain point <math>x_0</math> that has <math>G(x_0)>0</math>, or else <math>G(x)\equiv0</math>, can't be normalized. Then <math>G(x_1)+G(x_2)\ge2G(x_0)>0</math>. Now let <math>x_1\rightarrow-\infty</math>, <math>x_2\rightarrow+\infty</math>, in order to satisfy the above eqation <math>G(x_1\rightarrow-\infty)</math>,<math>G(x_2\rightarrow+\infty)</math> cannot be 0 at the same time, so the integral <math>\int |\psi(x)|^2\,dx=\int G(x)\,dx</math> will go to infinity, cannot be normalized.

is there any relation between Hermite polynomial and Legendre polynomial???

I do not quite understand why the <math>a_-\psi_n(x)</math> should be zero. Can it be some value between zero and <math>\frac{1}{2}\hbar\omega</math>?