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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?

I believe what he is trying to do here is say that the <math>\psi(\xi)</math> is equal to something “( )” times the exponent. Then he replaces the “( )” with <math>h(\xi)</math> and wants to solve for the <math>h(\xi)</math>, which he states is easier than solving for <math>\psi(\xi)</math> itself.

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>?

It has to be zero because of quantization of energy. If you apply the lowering operator <math>a_-</math> to a wavefunction repeatedly, you will travel down the “energy ladder” in discrete increments. The <math>a_-</math> operator is our method of “stepping down” a rung of energy. At some point, you have to reach a state of zero energy, since we can't have negative energy. The ground state <math>\psi_0</math> of a wavefunction is defined as the state with the smallest amount of energy a particle can possess in a binding potential (<math>\frac{\hbar\omega}{2}</math> for the simple harmonic potential). Since <math>\psi_0</math> is the lowest energy state with nonzero energy, applying the lowering operator <math>a_-</math> will step down a rung on the energy ladder to zero energy.

<math>\(a_{-}\)\psi_{0}\(x\)</math> must be non-normalizable (read: identically 0) or it wouldn't be <math>\psi_{0}</math> but <math>\psi_{1}</math> instead.

I don’t think about it in terms of energy. Thinking about it that way led me to the same question you asked. Here's how I think about now, after asking the Prof about it: We know that for any wavefunction there is some ground state, <math>\psi_0</math>, which is a stationary state; it’s the bottom rung on the ladder. Anywhere below this ground state the wavefunction cannot exist at all. So, if we step down one step from <math>\psi_0</math>, we’ll get zero (no wavefunction). That is, <math>a_-\psi_0(x) = 0</math>. Then we solve and magically get a value for <math>\psi_0</math> for the wavefunction of a harmonic oscillator, which is cool.

Would the Algebraic Method (with or without ladder operators) work with potentials other than the SHO? The spring potential is an interesting classical problem, but there are still so many more. Will some form of the Algebraic Method work in general? Or must other potentials be solved analytically?

With regards to the end of the analytical solution to the harmonic oscillator potential (p.54): I understand that the power series must terminate to prevent the asymptotic behavior as x goes to infinity. After this is where I get confused. We choose some arbitrary n so that the coefficients for all higher terms are 0. I see how K = 2n + 1 fits into this requirement from Equation 2.81. My question is, why can we decide that the power series terminates at arbitrary n for a corresponding energy <math>E_n</math>? I see how the quantization falls out of the math, but the logic of it all isn't clicking.

I noticed at the beginning of the power series method for the harmonic oscillator, Griffiths assumes that x is very large (bottom of p.51). But aren’t we usually looking at small x in a harmonic oscillator? How is Griffiths defining ‘very large x’ so that the solution is still useful and valid?

I'm also confused by this…Isn't x supposed to represent the position of the oscillator? Why are we only looking at large displacements?

It's time to move on to the next Q_A: Q_A_0925