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What sections is the quiz covering again?

I believe it is just up to and including section 2.3.

When using eq. 2.67 from the book, how exactly does the (a+)^n portion of the equation work? If I wanted to find <math>\psi_2(x)</math>, do I have to first square the a+ operator, or can I find <math>\psi_1(x)</math> and apply the operator again? Or does it not matter at all?

You can find <math>\psi_1</math> first and go from there. You'll just have to keep track of the <math>\frac{1}{\sqrt{n}}</math> terms.

you can do it both ways. you can operate on <math>\psi_0</math> once and get <math>\psi_1</math> and operate on <math>\psi_1</math> to get <math>\psi_2</math> or you can use (a+)^2 (which expands to give you 4 terms) and operate on <math>\psi_0</math>.

Back for another question. How do we find the probability of getting a specific energy (last problem of the practice quiz)?

The probability of finding the particle to have the energy corresponding to a specific state is <math>C_n^2</math>.

Check out the solutions to problem 2.13 part D. It is asking for the exact same thing.

Thank you. That was my first inclination, but I thought it was asking for something more complex.

Where is the square root of one comes from when calculating <x> in problem 2.13 in the solution?

I have the same question. My best guess is that it is using equation 2.67 with n=1. But somebody please correct me if I'm wrong!

Here is a study tip that I think might help with QM and I leave this tip up to discussion towards refinement. – Think of QM laterally in terms of big-concepts and sub-concepts. For example, delineate between the Time-Independent Schrodinger equation (big-concept) and separable solutions (sub-concept). The next part of the study tip is to think about the “How” and the “Why”. It's good to know how to solve separable solutions but it may be more important to know “Why” to use them. Focusing on the How and the Why of sub-concepts may make understanding the How and Why of big-concepts easier. This is my thought anyways. What do you think???

I like your idea, Green Suit, it's easiest for me to think of it that way, too. I'll get a bit more specific:

Start with the time-dependent Schrodinger equation: <math>-i\hbar\frac{\partial\Psi}{\partial x} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi</math>. Assuming that the wavefunction <math>\Psi</math> can be described by a product of two one-variable functions gives us separable solutions:

<math>\Psi(x,t) = \psi(x)\varphi(t)</math>

Plugging this product back into the time-dependent Schrodinger equation, we can solve two separate ordinary differential equations and obtain separable solutions. Next, we consider a number of time-independent potentials. This allows us to ignore the <math>\varphi(t)</math> part of the wavefunction since it will remain the same, and solve the Schrodinger equation independent of t.

The infinite square well, the simple harmonic potential, the finite square well, the free particle, and the Dirac-Delta potential are all examples of solving the time-independent Schrodinger equation for a different potential V.

Other random thoughts:

The general solution for any time-independent potential is a linear combination of all of the stationary states (<math>\phi_n(x)</math>'s - when you think about a linear comb of solutions, it makes sense only for time-dependent Schrodinger equation, not time-independent version because for the latter, each solution has unique and different <math>E_n's</math> on the RHS of the equation, so adding solutions with different <math>E_n's</math> won't satisfy the Schrodinger equation. i.e. linear combination of eigenfunctions are not an eigenfunction (of the Hamiltonian). Therefore, this should be <math>\Psi_n(x)</math>'s - Yuichi). You can also add on the time function and take a linear combination of all <math>\Psi_n(x)</math>'s to get the time-dependent solution.

Does anybody know what the operators for x and p are in terms of the raising and lowering operators? I didn't write it down and forgot what the constants in front were.

Eq. 2.69 in the book shows them as:

x = sqrt(h_b/2mw)(a_+ + a_-) & p = i sqrt(h_b mw/2)(a_+ - a_-)

Sorry, haven't had time to figure out math formatting yet. By the way how can I get this response to indent?

On the solution to discussion problem #3 the result for c_n says:

c_n = {0 if n is even} and {“whatever” if n is odd}

However the next line for c_2 (where n is even right?) has a non-zero answer…. What is going on here?

The answer should say Cn=0 for n even except for n=2.

Notice how the final eq. for Cn has an n^2-4 term in the denominator, so it doesn't work for n=2, since the denominator would be zero. To get Cn for n=2 you can plug n=2 into the original expression for Cn, which winds up being pretty straightforward since you get a sin squared term right away instead of sina * sinb. In fact, the second level, n=2, winds up having the largest contribution.

By the way, to indent your responses just use 3 equals signs on each side of your name-time-date instead of 4.

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