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Professor Chris Hammel, Ohio State University (OH), Physics, Quantum Mechanics, ProblemSet 2 Solution, Probability,Quantum Mechanics,un-normalized probability,normalized probability, spherical harmonic eigenfunctions,Infinite Well,Spherical Harmonics,Uncertainty,uncertainty principle,average momentum,Gaussian uncertainty principle,Operators,eigenvalue equations,Probability Density Current,Schrodinger's Equation,Symmetry,simple harmonic oscillator,Two-State Systems,LASER,Light Amplified Stimulate
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Quantum Mechanics }Probability The careless error here would be to just directly square the grids. When one remembers the significance of the meaning of the probability , one finds that one must square the wave function, and not the grids. The total probability is, The un-normalized probability from to is, The normalized probability is thus , as in choice (E).
Quantum Mechanics }Probability The careless error here would be to just directly square the grids. When one remembers the significance of the meaning of the probability , one finds that one must square the wave function, and not the grids. The total probability is, The un-normalized probability from to is, The normalized probability is thus , as in choice (E).
Quantum Mechanics }Probability Recall that Given the wave function in terms of the spherical harmonic eigenfunctions, one has it totally
Quantum Mechanics }Uncertainty One can make a good stab at this problem by applying the uncertainty principle. I. If the average momentum of the packet is 0, then one violates the uncertainty principle. See IV. II. Maybe. III. Maybe. IV. True, recall the Gaussian uncertainty principle. Since I is false, choices (A), (C), and (E) are out. Choices (B) and (D) remain. Take the conservative approach and choose (B).
Quantum Mechanics }Operators This problem can be solved (without much knowledge of quantum mechanics) by noting the following general arithmetic trick:. The problem gives the Hamiltonian , which has the same form as the arithmetic trick above. Thus,. Recalling some basic linear algebra, one can make use of the eigenvalue equations supplied with the problem defining the eigenvalues of the wanted operators,. Thus, , where one has applied the eigenvalue equation above and generalized it for the case
. From a bit of math manipulation, one has arrived at choice (D).
Generalizing the idea of a current from classical physics to the idea of a probability current, one takes the time derivative of the probability to get , where the product-rule for baby-math derivatives has been used and the derivative has been taken inside the integral because the integral and derivative are with respect to different variables. Plugging in the expression for from the Schrodinger's Equation, one gets , where the terms involving V's cancel out, and thus, Rewriting , one can eliminate the integral in the probability current by applying the fundamental theorem of calculus (to wit: ),
. But, since the probability current is usually define as , one has . (Aside:) One can print-out a cool poster or decent T-shirt iron-on to remember the Schrodinger's Equation (among other miscellanai) at a site the current author made several years ago, \begin{quote} http://anequationisforever.com/ds.php \end{quote} One can remember the general form of the probability current by recalling that it has to do with the difference of times its conjugate. Right, so onwards with the problem: The problem gives the wave function, so one needs just chunk out the math to arrive at the final answer, Thus, and, , where one notes that the imaginary terms go to unity from the complex conjugate. Plugging this into the probability current, one arrives at the expression for choice (E).
Quantum Mechanics }Symmetry One recalls the simple harmonic oscillator wave functions to be symmetric about the vertical- axis (even) for the 0th energy level, symmetric about the origin (odd) for the first energy level, and so on. If there is a wall in the middle of the well, then all the 0th energy level wave function would disappear, as would all even wave functions. Recall the formula for SHO. The first few odd states (the ones that remain) are , etc. This is choice (D).
