Thursday, December 14, 2006
I survived quantum physics
Yesterday I took the final in quantum physics, which actually wasn't too difficult. (The final, anyway.) I didn't think we'd get to perturbation theory or calculating the average energy of a superposition of eigenstates, but we did.
The energy of an eigenstate psi(n) is (pi^2)(n^2)(hbar^2)/(2mL^2). Each eigenfunction has a (complex) coefficient, and the expectation value of the energy of a wavefunction is equal to (the sum from 1 to n) of (C*C)E(n). C*C is, of course, the magnitude of C^2. The wavefunction 2psi(1) -5psi(2) + 7psi(3) (note that this is not normalized) has the average energy 4(1)+25(4)+49(9) = 545 times E(1).
The equation used in introductory perturbation theory is too complicated to write here without MathML (since it uses bra-ket notation) so I won't include it here in its proper form. (My blog entries are imported onto another site, which is where this blog gets many of its readers). For the mathematically-inclined reader, if we denote a solution c proportional to the integral from 0 to t of dt' times e^-i[E(k)-E(m)]t/hbar times the bra-ket containing psi(k), a z-term, and psi(m), the maximum value of this integral will occur when E(k) - E(m) = hbar*omega. This gives e^-i(omega)t, which is a standard, exponentially-decaying time part of psi.
The exam covered optical transitions, but in an unusual form. An optical transition occurs when l (the angular momentum quantum number) and m(l) (the azimuthal quantum number) change a certain amount. If they do not change by a specific integer amount (for l it is +- 1 and for m(l) it is 0 or +-1), then the probability of an optical transition occurring is zero.
But...
An optical transition may still be observed, at least according to chemists. They have detected these optical transitions, which may signify that there is a flaw in the physicists' integrals of the theta- and phi-dependent parts of the wavefunction.
I've never heard anyone speculate about this; normally I just hear about attempts to prove the theory of relativity wrong. Now what about the state of relativistic quantum mechanics...? ¶ 12:15 AM
The energy of an eigenstate psi(n) is (pi^2)(n^2)(hbar^2)/(2mL^2). Each eigenfunction has a (complex) coefficient, and the expectation value of the energy of a wavefunction is equal to (the sum from 1 to n) of (C*C)E(n). C*C is, of course, the magnitude of C^2. The wavefunction 2psi(1) -5psi(2) + 7psi(3) (note that this is not normalized) has the average energy 4(1)+25(4)+49(9) = 545 times E(1).
The equation used in introductory perturbation theory is too complicated to write here without MathML (since it uses bra-ket notation) so I won't include it here in its proper form. (My blog entries are imported onto another site, which is where this blog gets many of its readers). For the mathematically-inclined reader, if we denote a solution c proportional to the integral from 0 to t of dt' times e^-i[E(k)-E(m)]t/hbar times the bra-ket containing psi(k), a z-term, and psi(m), the maximum value of this integral will occur when E(k) - E(m) = hbar*omega. This gives e^-i(omega)t, which is a standard, exponentially-decaying time part of psi.
The exam covered optical transitions, but in an unusual form. An optical transition occurs when l (the angular momentum quantum number) and m(l) (the azimuthal quantum number) change a certain amount. If they do not change by a specific integer amount (for l it is +- 1 and for m(l) it is 0 or +-1), then the probability of an optical transition occurring is zero.
But...
An optical transition may still be observed, at least according to chemists. They have detected these optical transitions, which may signify that there is a flaw in the physicists' integrals of the theta- and phi-dependent parts of the wavefunction.
I've never heard anyone speculate about this; normally I just hear about attempts to prove the theory of relativity wrong. Now what about the state of relativistic quantum mechanics...? ¶ 12:15 AM
