Recent content by EE18

Perhaps I don't follow, but isn't your last line more or less what I've given in my last line? Would you be able to comment on how ##Q^2## now provides a measurement of the initial state of the object's position ##Q^2## beforehand?

$$\newcommand{\bra}[1]{\left \langle #1 \right \rvert} \newcommand{\braxket}[3]{\left \langle #1 \middle \rvert #2 \middle \rvert #3 \right \rangle} \newcommand{\ket}[1]{\left \rvert #1 \right \rangle} \newcommand{\expec}[1]{\langle #1 \rangle}$$ Ballentine asks us the question at the end of...

How come you believe this to be true? Even a slight hint would get me towards understanding how this construction actually means that $R$ as described is the ground state wavefunction. Can you explain why being negative should be a problem? Also, would you be able to make a comment about 2) at...

Ballentine, in his Chapter 8.1, appears to give the attached recipe for *in principle* preparing an (almost) arbitrary (pure) state (of a particle with no internal degrees of freedom) by the method of "waiting for decay to the energy ground state". My questions are fourfold: 1) From (8.1), we...

I've given the question as an image as some of the formatting was difficult for me in the small window given: My work is below. I got (a), but cannot get (b): (a) It was a theorem proved in the text that any measurement on one subsystem will always be fully determined by the reduced state...

I am struggling with the latter, and think that I somehow need to assume ##f## is real-valued to proceed? My work: The position distributions are equal since $$P_{-m}(\mathbf{x}) = |\Psi_{-m}(\mathbf{x})|^2 = |f(r)Y_l^{-m}(\theta,\phi)|^2 = |f(r)(-1)^m(Y_l^{m})^*|^2 =...

This was fascinating. Thank you to you and @TSny for this very interesting addendum!

Just doing this out for psterity. We have $$T\frac{d^2}{dT^2}\left(BT\right) = T\frac{d}{dT}\left(B'T + B\right) = T^2B'' + 2TB' = \frac{d}{dT}\left(T^2\frac{dB}{dT}\right).$$ In so doing, I suppose I've verified that ##u## as written represents an antiderivative of ##c_v## (with respect to...

As usual, it looks like I should have referred to the first edition! Thanks for the tip. In this case (all equations referred to are second edition), it appears that if one takes (13.29) as given, then one can argue that $$du = Tds - Pdv = c_vdT + T\frac{P}{T}dv -Pdv = c_vdT \implies u =...

I guess I'm not quite sure how you got to that form from the total differential you gave above. Did you integrate first at infinite volume (fixed) so that you had an ideal gas, and then proceed with the volume part of the integral? But then it seems like there's no denominator now?

You have ##c_v## in the expression too though? I think that's where I'm not quite following.

I'm not sure I see how the final expression gets me there. The form for ##C_v## is not immediately clear to me, and it seems like you're tacitly saying I could use that in your final expression?

In his Chapter 13.3 (2nd edition), Callen gives the standard form for the virial expansion for the mechanical equation of state of a fluid as an exapnsion in powers of the molar volume ##v##: $$P = \frac{RT}{v}\left(1 + \frac{B(T)}{v} + \frac{C(T)}{v^2} + \dots \right) \equiv P_{ideal} +...

Oof, of course. ##\exp(2+\ln K_s(T)) =e^2K_s## which is of course not what we have here. My bad, and thanks for the clarification on this silly error.

You are saying to use $$\ln K_d(T) = 2\ln K_s(T) = \ln K^2_s(T)\implies K_d = K_s^2$$ which makes sense to me. I'm embarrassed to say I don't know what I'm doing wrong by using $$\exp(\ln K_d(T)) = K_d = \exp(2\ln K_s(T)) = e^2 exp(\ln K_s(T)) = e^2K_s$$ which is different. What elementary...