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Let's consider the Q in footnote 22 (see image). Q=##\frac{1}{2}##mω2x2, where ω is a function of t. Then is ∂Q/∂t = mω##\frac{dω}{dt}##x2? Even though for simple harmonic motion, x is also a function of t (ie. x=x0##\sin##(ωt), where x0 is the amplitude), we do not care how x depends on t when...

∂Q/∂t refers to the partial derivative, not the total derivative. The book goes on to talk about operators that explicitly depend on time (see attached images). In doing the partial derivative ∂Q/∂t, we do not care about how x depends on t, but only how Q explicitly depends on t. But no matter...

Suppose Q=2x+t and x=t2, then ∂Q/∂t=1. But Q can also be written as Q=x+t2+t, then ∂Q/∂t=2t+1. We now have 2 different answers. But I think there can only be one correct answer. In reference to the equation in the image, no matter we write Q=2x+t or Q=x+t2+t, <Q> should be the same, so the LHS...

Yes I meant the response "There is currently neither conclusive evidence for the multi-histories interpretation nor for the single-history interpretation. We simply do not know which is true at this point in time." is an impartial, fair response in relation to EPR's response "What evidence is...

Similarly, someone could say: "Currently, what conclusive evidence is there of a one-and-only history in the macro world? There is none. It's not an opinion. It's a scientific fact." So an impartial, fair response would be: "There is currently neither conclusive evidence for the multi-histories...

I realize it's not easy to get neutral, impartial responses. Most of us have our own perspectives, beliefs and biases, and we allow these biases to influence our response, knowingly or unknowingly. My attempt of an impartial, fair response would be: "Feynman's sum over histories is a...

Feynman's method shows that the historical path of a particle is not encrypted into the present state of the particle, so this is also true for a collection of particles, this implies that our history is also not encrypted into our present state?

This, at best, says Feynman's sum over histories is consistent with the orthodox, one-history Copenhagen-style interpretation? But Feynman's sum over histories could also be consistent with a multi-histories interpretation? Before Feynman's sum over histories, most of us were oblivious to or...

What does Feynman's sum over histories mean to the interpretation of our world? Does it mean that we (or a particle) do not have a definite history, but only the most probable one?

Does the sign always say "the cat is alive", even if the cat is found dead when the box is opened? And why must the sign be inside the box? It seems the sign just has a certain probability of being true, say 60% true and 40% false. But is that the same as being in a superposition of being true...

If a cat can be in a superposition of being dead and alive, why can't a statement be in a superposition of being true and false?

So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too? So its radius of curvature ##r=\frac{c^2}{9.81}\approx10^{16}## m? But what if it travels straight down towards the Earth? It cannot move faster than ##c##, so its acceleration would have to be zero. So given...

Could you explain further? If light is massless, then its motion should be unaffected by Newtonian gravity.

Gravity can be described not as a force but a curvature of spacetime. I assume this can’t be done to the other 3 fundamental forces. If so, then we cannot treat gravity in a way similar to the other forces. Why then does QFT postulate the existence of gravitons? Why does it attempt to treat...

My conclusion is because I use ##p^4=p^2p^2## to arrive at the error expression for ##p^4##, I cannot later use that expression to argue or conclude that ##p^4\neq p^2p^2##. The two terms in the middle simplifies to ##r^2(2-\frac{r}{a})e^{-3r/2a}##, ignoring any constant factor. I used...

This is weird because I get the error expression for ##p^4## in post #7 by assuming ##p^4=(p^2)^2##. Ignoring the constant factor ##-4\pi\hbar##, ##\langle f|p^2(p^2g) \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\langle p^2f|p^2g...

For ##g=\psi_{100}## and ##f=\psi_{200}##, ##\left.\left( r^2 f\dfrac{dg}{dr} - r^2g\dfrac{df}{dr} \right)\right|_0^\infty=\left.\left(\frac{1}{8\sqrt{2}\pi a^5}r^3e^{-3r/2a}\right)\right|_0^\infty=0##, where ##a## is a constant.

Does this mean that a measurement of ##p^2## is only approximately real valued? What does that even mean?

##p^2## is hermitian then. The contradiction still persists.

All operators for observables must be hermitian. If ##\hat{p}^4## is not hermitian, then what would you obtain when you measure ##p^4## or ##E^2##? Would you get complex-valued measurements? What would it mean?

c(r) has the factor ##r^2##, so it is exactly 0 at r=0.

Everything involved is always differentiable. Taking ##f=\psi_{200}=(2-\frac{r}{a})e^{-r/2a}## and ignoring all constant factors, $$r^2\frac{df}{dr}=(4r^2-\frac{r^3}{a})e^{-r/2a}$$ $$\frac{1}{r^2}\frac{d}{dr}r^2\frac{df}{dr}=(\frac{r}{a}-\frac{2}{a}+\frac{16}{r})e^{-r/2a}$$

Yes the boundary term does not vanish, indeed. But how can this be? All the hydrogen radial wave functions are infinitely differentiable. Doesn't it contradict the theorem?

It is, because no matter how many times you differentiate ##e^{-r}##, multiplying and dividing by ##r^2##, in whatever order of these 3 operations, the result is still proportional to ##e^{-r}##.

Yes they are since hydrogen radial wave functions all have the ##e^{-r}## term.

##p^2## is hermitian does indeed hold for all states, including ##l=0##, but not ##p^4##.

##p## is the momentum operator as seen from [6.52]. ##p^4=(p^2)^2## and ##p^2## is hermitian as seen from [6.51].

He is saying ##p^2## is hermitian but ##p^4## is not for the same pair of states, f and g, both with ##l=0##.

Intro to QM, David Griffiths, p269

Why is ##p^4## not hermitian for hydrogen states with ##l=0## when ##p^2## is? Doesn't this contradict the following theorem?

I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature. If there are two kinds of measurements—one, quantum and the other, classical—then how do you tell them apart properly?

I guess you meant this part: All these I understand, but it does not mention how the condition ##\psi(x+a)=e^{iKa}\psi(x)## is motivated. Be aware that it is only by first assuming this condition to be true that we could get ##\psi(\vec{x}+N a)=e^{iKNa}\psi(\vec{x})##. If we start by not...

So is it just pure coincidence that the double-slit interference for light and electrons are described by the same equation as the classical water-wave interference?

Could you explain quantum coherence in simple terms? A water wave of wavelength ##\lambda## should behave as a particle with momentum ##p=\frac{h}{\lambda}##. How can we observe this particle? And how can we collapse the wave function of this particle so that it passes through either the left...

Electrons passing through a double slit is in a superposition of passing through the left slit and the right slit, thereby producing an interference pattern on the screen. But when a detector is placed to detect which slit the electrons pass through, the interference pattern is destroyed. How...

Since you understood the argument with the Wronskian, could you explain to PeterDonis that it doesn't impose the following condition on ##K##: ##e^{i K a} = e^{- i K a}##? (Note that for the case of the free particle, K=k. K is defined by the Bloch's condition [5.49], while k is defined by the...

It is not a stationary state. But this is not a counter example because [5.48] is the time-independent Schrodinger solution. There exists an allowed value of k such that a solution to the time-independent Schrodinger equation is not a solution to the Bloch's condition. Could you explain how the...

There exists an allowed value of k such that a solution to the time-independent Schrodinger equation [5.48] is not a solution to the Bloch's condition [5.49], but Griffiths said (according to @PeterDonis) this is impossible: all solutions to the Schrodinger equation is a solution to the Bloch's...

##\psi_{\lambda_1}## and ##\psi_{\lambda_2}## have the same E but different ##\lambda##, same eigenvalue for H but different ones for D! (Different ones in general: ##\lambda=\pm1## are the exceptions.) This is proven by B&J: YES! YES! YES! This is what I have been saying several times in...

Ok, let's hear the opinion of more people first. [5.48] is Schrodinger equation. [5.49] is Bloch's theorem or Bloch's condition. [5.62] is in post #65: You are right that we need to take into account the Dirac-delta potential at ##x=0## and ##x=a##. But it turns out that [5.59] is still the...

Does the principle of conservation of energy (PCoE) rule out the possibility of creating such an entangled pair of particles A and B (where now neither A and B is taken to be the rest of the Universe, to avoid the difficulty in dealing with the rest of the Universe)? Is it the case that (the...

[5.62] is used to search for common eigenstates of H and D, since it is obtained from the eigenvalue equations of H [5.48] and of D [5.49]. Therefore, not any A and B would satisfy e.g. [5.62]. Isn't it true that any A and B would satisfy [5.48] for a particular E since [5.59] is its general...

Is QM consistent with the assertion that if a particle A is newly made to be in the superposed state ##\psi_A=c_1\psi_{A, E_1}+c_2\psi_{A, E_2}##, then the rest of the Universe (apart from A) can be prescribed the state ##\psi=c_2\psi_{E_3}+c_1\psi_{E_4}##, where ##E_1+E_3=E_2+E_4##? In other...

How do we know that this is not the case? Particle B could be taken as the rest of the Universe apart from the newly created particle A.

What you said seems to contradict Griffiths's statement below: Griffiths said states with energies in the gap region are physically impossible, but you said these states are physically relevant. A definition of "growing mode" and "decaying mode" would be helpful, and also a concise...

Ok I didn't show an example explicitly as I thought B&J's paragraph following (4.190) is an adequate proof of what I have been saying. Anyway, you can find one such example from Griffiths: For each value of K in [5.56], we solve [5.64] graphically. For each K, we draw a horizontal line in...

@PeterDonis Do you agree that there exists an allowed value of k such that ##\psi=\sin(kx)## is an eigenstate of H, but not of D?

This is because

This is false. Consider the simplest case of ##\psi=\sin(kx)##. It is an eigenstate of H, but not of D. Even with complex eigenvalues, you cannot write ##\sin(kx)## as a complex multiple of ##e^{ikx}## alone or as a complex multiple of ##e^{-ikx}## alone.