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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???

No, although they both are recursive relations of sorts, they aren't related in any way. But, Hermite Polynomials are special cases of Laguerre polynomials, if your interested in looking into that.

I do not quite understand why the <math>a_-\psi_0(x)</math> should be zero. Can it be some value between zero and <math>\frac{1}{2}\hbar\omega</math>?