Quantum_solution_a_la_jeopardy {sin(x),sin(ax)}

Slide 25a: The Jeopardy Approach to Quantum Mechanics

To better understand the ideas between Quantum Mechanics, one can take the “Jeopardy Approach” and use the answer to find the question that was asked. In this case, the answer is the wave function’s equation and the original problem is the relationship between an electron’s kinetic energy and its mass and potential energy(x). “Potential energy(x)” means the potential energy in terms of x.

We now know that Kinetic Energy’s actual formula is (-C/m)*(curvature of psi)/psi. Using this, we begin to work backwards.

Let’s say psi = sin(x). What would be the Kinetic Energy? Well, using the formula we can deduce that KE is (a negative constant) divided by (a mass) times (the second derivative of sine, which represents its curvature) divided by (psi, which is sine). The second derivative of sin(x) is -sin(x). This means the sine functions will cancel because one is in the numerator and the other is in the denominator. The two negative signs also cancel for the same reason. This gives us: C/m as the KE. What does this mean? “X” doesn’t matter! Therefore, PE must remain constant. This particle must be floating in free space.

Let’s try another one: psi = sin(ax). The second derivative of this is -a^2*sin(ax). This, again, represents the curvature of the graph at point x. When the curvature is divided by psi, the sin(ax)’s will cancel! This leaves you with -C/m * -a^2 as the KE. The negatives undo each other, and the KE can be rewritten as: a^2*C/m. “X” is once again not a factor in the KE! However, the KE will increase in proportion to a-squared.

-Andrew Williams

If a is greater than 1, psi = sin(ax) has a shorter wavelength than psi = sin(x) but the KE corresponding to the former is greater by a factor of a2. We can deduce that wavelength squared is inversely related to kinetic energy.

-Yanyao Fu

Obviously it wouldn't be a problem if we used Ψ=cos(x) instead of Ψ=sin(x): the second derivative of cos(x) is -cos(x) and the cosines cancel to give the exact same kinetic energy expression (C/m) as with sin(x), so the kinetic energy would be the same for using sine and cosine. The fact that they are not the same function doesn't create a problem because the cosine function is a mere horizontal shift of the sine function, and we already know from the given infomation that potential energy doesn't change based on position, and from our calculations we know that kinetic energy doesn't change based on position, either. So, at least for particles in free space, if we can use sin(x) as our Ψ, then we can also use cos(x).

The problem, however, happens when we use Ψ=sin(ax) instead of Ψ=sin(x). In this case, the kinetic energy, while independent of x, varies based on the value of a we choose. Since we obtain an expression for kinetic energy from our Ψ function, then aren't we arbitrarily determining the kinetic energy when we choose our value of a? According to the criteria so far described in the powerpoint, we can choose any finite Ψ function such that kinetic, and hence potential energy are independent of position (at least for a particle in free space). But it concerns me that we can choose any kinetic energy we want, as long as we satisfy that sole condition for Ψ. Can we do that, or is there another factor we have to consider when choosing an appropriate Ψ for a particle? Do we have to worry about what the kinetic energy actually is? (in classical physics you don't worry about the fact that a ball can have any velocity with its associated kinetic energy - jmm)

(I assume these questions will be answered very soon in the course. If they will be, then they are a good indicator of what we are supposed to know as of this particular slide.

I got the Ψ symbol in here by going into Microsoft Word, choosing "insert symbol" and inserting the Ψ symbol into a blank document, and then copying and pasting the symbol from the Word document into the editing screen for the wiki. If anyone knows a bettter way, please tell me.)

-Rick Russotto

An interesting thing to note is the negative value for kinetic energy when Ψ= e^x and Ψ=e^-x is picked as a solution. Although the curvature of the wave function and the wave function itself simplify nicely, we are still left this negative energy. This obviously clashes with the more classical approach to solving for kinetic energy (K=.5mv^2). With this formula, a negative product would be impossible; mass is positive, and the velocity term is squared. However, in the Schrodinger formula for kinetic energy a negative outcome is permissible. Indeed, as the slide describes, it is more than just a "mathematical curiosity". This actually happens to all electrons that are bound to the nucleus, especially at large values of r when (1/r) will stop changing significantly.

-Tyler Elkington

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