Recent content by cattlecattle

sorry, I failed to see which step was not applicable due to unboundedness

yes - because the expected value of A-B will be zero for all psi, then by taking psi to be A-B's eigenvector, you reached the conclusion that A-B has only zero eigenvalues, which means A-B must be a zero operator.

I'm studying Shankar's book (2nd edition), and I came across his equation (15.3.11) about spherical tensor operators: [J_\pm, T_k^q]=\pm \hbar\sqrt{(k\mp q)(k\pm q+1)}T_k^{q\pm 1} I tried to derive this using his hint from Ex 15.3.2, but the result I got doesn't have the overall \pm sign on the...

It's the |\Psi|^2, a real number

the reason your orthogonality yields delta function relies on the integration domain being the entire real line, which is not true in bound states, in fact bound states all have discrete spectrum, and scattering states continuous spectrum

No, the general result is if \Omega|\omega\rangle=\omega|\omega\rangle then \langle\omega|\Omega^\dagger=\langle\omega|\omega^* There is usually no relation between \Omega|\omega\rangle and \langle\omega|\Omega

\langle \omega_2 | \Lambda \Omega | \omega_1\rangle =\langle \omega_2|\Lambda \omega_1|\omega_1\rangle = \omega_1\langle \omega_2|\Lambda|\omega_1\rangle\\ \langle \omega_2 | \Omega\Lambda|\omega_1\rangle = \langle\omega_2|\omega_2\Lambda|\omega_1\rangle=...

It follows when the time interval is infinitesimal, where the velosity is considered constant (xb-xa)/(tb-ta) during the integration interval.

Almost all proofs of 0=1 involve symbolic manipulation of infinity. This is no exception. Intuitively, when in eigenstate of P, particle is in definite momentum, but <p|X|p>, the value of position is totally indefinite, so intuitively it's not well defined. To see this more rigorously, consider...

At your second equal sign, on the left, you have an expected value (complex number), on the right, you have a ket.

It has the same statistical meaning as usual variance. The deviation of any particular measurement from the expected value is A-<A>, the average value of this quantity is by definition zero. so in order to define variance, we square it (so it's no longer zero), and take the average. (followed by...

It's just a reasonable postulate. I couldn't imagine anything physical to be complex number, if you can, please give an example (even hypothetical ones). I would personally visualize physical attributes to be something you can read from an equipment. What does it even mean if you say a physical...

This concept is introduced to distinguish eigenstates of the same eigenvalue from each other, i.e., degenerate states. For operator A, if it has degenerate eigenvalues, say a, all the eigenvector for 'a' form a subspace of dimension > 1, so it's not possible to say for sure what state the...

My proof above claims further that the energy of any eigenstate is no less than \hbar\omega/2.

It can be shown that any eigenstate of the QHO has positive energy. To see this, first we show the expectation value of the energy measurement for any state is positive. \langle H \rangle = \langle m\omega^2X^2/2 + P^2/2m\rangle = \frac{m\omega^2}{2}\langle X^2\rangle + \frac{1}{2m}\langle P^2...