Relativity, time, and quantum mechanics

[anuttarasammyak]

[Mentors’ note: Spun off from this thread]
I understand that you are interested in the concept of time in quantum mechanics. Here are examples of my questions, in case they may be of reference.

１.How can one calculate the proper time of an electron in the 1s orbital of a hydrogen atom? How should the Lorentz factor be applied to motion that is not classical?　Would that result in a superposition of different proper times?

2. To determine the invariant spacetime interval between events associated with a particle, one would obtain the spacetime coordinates (t, x, y, z). However, due to the uncertainty principle, measuring the spatial coordinates x, y, z imparts momentum to the particle, so its rest frame changes from the original one. It would be a mess. Since there is no measurement operator for t, does time play the role of reconciling this inconsistency?

3. It appears that, with respect to a measurement at time zero, when collapse takes place in interaction with apparatus or environment, the state before measurement and the state after measurement are asymmetric, e.g. entangled and disentangled. Could this be the origin of thermodynamic or arrow of time?

 

[Nugatory]

１.How can one calculate the proper time of an electron in the 1s orbital of a hydrogen atom?

How should the Lorentz factor be applied to motion that is not classical?[/quote]We don’t. Proper time is the length of a path through spacetime between two events; the electron has no path or position so the notion “the proper time of an electron” is meaningless. Instead we work with the time measured by our lab clock between observation events.

2. To determine the invariant spacetime interval between events associated with a particle, one would obtain the spacetime coordinates (t, x, y, z). However, due to the uncertainty principle, measuring the spatial coordinates x, y, z imparts momentum to the particle, so its rest frame changes from the original one.

X,y,z,t are the coordinates of the observation event using the frame in which the lab is at rest. The “rest frame of the electron” never enters into the calculation (unsurprisingly, because it’s not defined).

3. It appears that, with respect to a measurement at time zero, when collapse takes place in interaction with apparatus or environment, the state before measurement and the state after measurement are asymmetric, e.g. entangled and disentangled.

The global collapse interpretation you are trying to use here is non-relativistic so cannot be applicable here. Instead we have to use the methods of quantum field theory, in which there is no global collapse.

Could this be the origin of thermodynamic or arrow of time?

We’ll find that in statistical mechanics.

 

[anuttarasammyak]

Addeentum to 1 of post #1 on proper time in quantum mechanics :
Consider a double-slit experiment with an electron beam at fixed accelerating voltage. Suppose that each electron carries its own clock and reads the time when it is just before the slits and again when it hits the screen. The difference in the readings would reveal which of the two slits the electron passed through. It therefore seems that precice proper time reading (= which-path information) cannot be compatible with the presence of interference.

 

[anuttarasammyak]

the electron has no path or position so the notion “the proper time of an electron” is meaningless.

Let me confirm for my understanding. Even in the 1s state of exotic hydrogen atom consisting of proton and muon, does muon not collapse due to its no proper time? If it collapses, won't the long-life effect of motion take place ?

 

[renormalize]

How should the Lorentz factor be applied to motion that is not classical?

We don’t. Proper time is the length of a path through spacetime between two events; the electron has no path or position so the notion “the proper time of an electron” is meaningless.

Let me confirm for my understanding. Even in the 1s state of exotic hydrogen atom consisting of proton and muon, does muon not collapse due to its no proper time? If it collapses, won't the long-life effect of motion take place ?

The answer by @Nugatory is incorrect. As suggested by @anuttarasammyak, experiments with muonic atoms show that bound particles do experience time dilation, as reported in this paper:
Relativistic time dilatation of bound muons and the Lorentz invariance of charge
Abstract: The relativistic lengthening of the lifetime of a decaying elementary particle bound in a stationary state of an exotic atom provides evidence that the particle is actually in motion even though such motion cannot be visualized classically. This application of special relativity to the particles within an atom helps illuminate an argument frequently used for the Lorentz invariance of electric charge.
I don't have access to this paper, but an answer at https://physics.stackexchange.com/questions/730311/ by someone who has read it states:
Muonic atom decay lifetimes were measured back in the 1960s, and these lifetimes agreed with relativistic theoretical calculations. For example, the theoretically expected time dilation for lead was 16-20%, and the experimental value was 14±4%. Not a very precise test, but enough to strongly suggest the atomic muons are subject to time dilation.

 

[Nugatory]

The answer by @Nugatory is incorrect. As suggested by @anuttarasammyak, experiments with muonic atoms show that bound particles do experience time dilation, as reported in this paper:

I think you misread what I said.

 

[renormalize]

I think you misread what I said.

That's certainly possible! Given that you say "the electron has no path or position so the notion “the proper time of an electron” is meaningless" how then would you interpret those experiments that purport to demonstrate decay-time dilation for muons in atomic stationary states?

 

[anuttarasammyak]

X,y,z,t are the coordinates of the observation event using the frame in which the lab is at rest. The “rest frame of the electron” never enters into the calculation (unsurprisingly, because it’s not defined).

Thank you for your reply. Point 2 of post #1 was pointless, so I withdraw it.

 

[PeterDonis]

@anuttarasammyak you are trying to analyze quantum experiments using relativity, but using non-relativistic QM. That's not going to work. If you want to try to analyze quantum experiments using relativity, you need to be using quantum field theory--in which many of the questions you're asking aren't even well posed.

 

[Nugatory]

That's certainly possible! Given that you say "the electron has no path or position so the notion “the proper time of an electron” is meaningless" how then would you interpret those experiments that purport to demonstrate decay-time dilation for muons in atomic stationary states?

No position means no path along which we can integrate to find the proper time. Instead we use QFT and calculate an amplitude for a decay event at various points in spacetime - "experiences time dilation" is not how I would describe that model. And note that the abstract of the paper does not use that term - it makes the interpretational(!) claim that the particle moves even though we cannot observe the motion.

 

[renormalize]

No position means no path along which we can integrate to find the proper time. Instead we use QFT and calculate an amplitude for a decay event at various points in spacetime - "experiences time dilation" is not how I would describe that model. And note that the abstract of the paper does not use that term - it makes the interpretational(!) claim that the particle moves even though we cannot observe the motion.

OK, I want to see if we're really on the same page, rather than in completely different chapters!
Let's consider a non-relativistic electron occupying the $n^{\text{th}}$ energy-eigenstate of a hydrogen-like atom, i.e., it's in a stationary "orbital" with energy $E_{n}=T_{n}+V\left(r\right)$, where $T_{n}\equiv p_{n}^{2}/\left(2m\right),V\left(r\right)\equiv-Ze^{2}/r$. Applying the quantum virial theorem to this central potential yields:$$2⟨T_{n}⟩=-⟨V⟩\Rightarrow\frac{⟨m^{2}v_{n}^{2}⟩}{m}=⟨\frac{Ze^{2}}{r}⟩\Rightarrow m⟨v_{n}^{2}⟩=\frac{Z^{2}me^{4}}{n^{2}\hslash^{2}}$$or $$⟨v_{n}^{2}⟩=\frac{Z^{2}e^{4}}{n^{2}\hslash^{2}}=\frac{Z^{2}\alpha^{2}c^{2}}{n^{2}}$$It then seems reasonable to define the average speed $s_n$ of the electron to be:$$s_{n}\equiv\sqrt{⟨v_{n}^{2}⟩}=\frac{Z\alpha c}{n}$$Since this quantity is non-zero for any orbital, to me it's completely natural to conclude that:

Although an electron bound in the $n^{\text{th}}$ stationary orbital of a hydrogen-like atom does not move along a classical path, it is nevertheless constantly in motion at an average speed of $Z\alpha c/n\,$.

Are you in agreement with this statement, or do you favor a different interpretation of the speed $s_n$ that somehow doesn't involve "motion" of the electron?

 

[PeterDonis]

Applying the quantum virial theorem to this central potential yields

Where did the $v_n$ in this formula come from? There was no $v_n$ in the previous formula for $T_n$.

 

[renormalize]

Where did the $v_n$ in this formula come from? There was no $v_n$ in the previous formula for $T_n$.

Huh? Doesn't $p_n^2=(m\vec{v}_n)\cdot(m\vec{v}_n)=m^2 v_n^2\,$?

 

[PeterDonis]

Huh? Doesn't $p_n^2=(m\vec{v}_n)\cdot(m\vec{v}_n)=m^2 v_n^2\,$?

For a free particle, yes. But you can't just assume that for a particle in an external potential. See, for example, the discussion in Ballentine, Section 3.4.

 

[anuttarasammyak]

For a free particle, yes. But you can't just assume that for a particle in an external potential. See, for example, the discussion in Ballentine, Section 3.4.

$$<T_n>=\frac{1}{2}m<v_n^2>$$ is also only for free particle ?

 

[renormalize]

For a free particle, yes. But you can't just assume that for a particle in an external potential. See, for example, the discussion in Ballentine, Section 3.4.

True, but per Ballentine's treatment of the hydrogen atom in §10.2:

Setting $M_e=m$ and taking the mass of the proton $M_p\rightarrow\infty$ turns Ballentine's Hamiltonian above (10.20) into $P_{e}{}^{2}/\left(2m\right)-e^{2}/\left|\mathbf{Q}_{r}\right|\,$, which is exactly the same form (with $Z=1$) that I wrote for $E_n$ in post #11. So with the caveat that the mass of the nucleus must be very large compared to that of the electron, I stand by my virial-theorem-based derivation in that post, as well as the conclusion that the electron is in motion with an average speed of $Z\alpha c/n\,$.

 

[anuttarasammyak]

Let me confirm for my understanding. Even in the 1s state of exotic hydrogen atom consisting of proton and muon, does muon not collapse due to its no proper time? If it collapses, won't the long-life effect of motion take place ?

I have read that muon could be captured by nucleus which shortens its life time. This model seems not appropriate to investigate the proper time and time dilation. Another toy model, e.g., muon in square well potential, might be considered instead.

 

[renormalize]

I have read that muon could be captured by nucleus which shortens its life time. This model seems not appropriate to investigate the proper time and time dilation.

The theoretical contribution of nuclear capture can be calculated and subtracted from measurements of muonic-atom lifetime in order to isolate the effect of time dilation.

 

[Nugatory]

It then seems reasonable to define the average speed $s_n$ of the electron to be….

You have a calculated a quantity that has the dimensions of distance over time, and you can interpret it as a speed (it's an interpretation because it is not observable even in principle and because there's no position that is changing with time) if that interpretation is helpful.

But it's a stretch to get from there to the proposition that a bound particle "experiences time dilation" (what could that phrase possibly mean? Even with macroscopic objects where there's a clean worldline along which proper time is measured, and even with a person following that worldline so that we can ask about their "experience", time dilation isn't something "experienced").

I'm not seeing how this discussion does anything for the confusion in the original posts.

 

[PeterDonis]

$$<T_n>=\frac{1}{2}m<v_n^2>$$ is also only for free particle ?

That's the only case for which you can take that as a unique result, yes.

 

[PeterDonis]

True, but per Ballentine's treatment of the hydrogen atom in §10.2:

All that shows is that the Hamiltonian has the form you wrote down. But my issue has nothing to do with that Hamiltonian not being correct for the hydrogen atom; I agree that it is, and I never said otherwise. You aren't addressing the actual issue I raised at all.

 

[renormalize]

You aren't addressing the actual issue I raised at all.

That's because I don't understand what your issue is.
In §3.4 Ballentine states:

The Hamiltonian (3.60) pertinent to the stationary orbitals of hydrogen clearly has $\mathbf{A}\left(\mathbf{Q}\right)=0$, so from (3.59):$$\mathbf{V}=\frac{\mathbf{P}}{M}\Rightarrow\mathbf{v}^{2}\equiv\mathbf{V}\cdot\mathbf{V}=\frac{\mathbf{P}\cdot\mathbf{P}}{M^{2}}\equiv\frac{\mathbf{p}^{2}}{M^{2}}=\frac{2M\mathbf{T}}{M^{2}}=\frac{2}{M}\left(E\,\mathbf{I}-\mathbf{W}\left(\mathbf{Q}\right)\right)$$from which it follows that:$$\langle\mathbf{v}^{2}\rangle=\frac{2}{M}\left(E-\langle\mathbf{W}\left(\mathbf{Q}\right)\rangle\right)=\frac{2}{M}\left(-\frac{Z^{2}Me^{4}}{2n^{2}\hslash^{2}}+\frac{Z^{2}Me^{4}}{n^{2}\hslash^{2}}\right)=\frac{Z^{2}e^{4}}{n^{2}\hslash^{2}}=\frac{Z^{2}\alpha^{2}c^{2}}{n^{2}}$$as I claimed in post #11.
If you still take issue with this result, I ask that you please describe your specific objection.

 

[renormalize]

You have a calculated a quantity that has the dimensions of distance over time, and you can interpret it as a speed (it's an interpretation because it is not observable even in principle and because there's no position that is changing with time) if that interpretation is helpful.

As displayed in post #22, Ballentine explicitly defines the velocity operator $\mathbf{V}$:

If "it is not observable even in principle and because there's no position that is changing with time" why does he introduce $\mathbf{V}$ in a textbook on quantum mechanics? Moreover, doesn't every Hermitian operator like $\mathbf{V}$ correspond to a physical observable?

 

[PeterDonis]

That's because I don't understand what your issue is.

You do now, at least one part of it.

The Hamiltonian (3.60) pertinent to the stationary orbitals of hydrogen clearly has $\mathbf{A}\left(\mathbf{Q}\right)=0$

Yes, because it's a Coulomb potential, which is purely scalar.

But there's another part as well:

doesn't every hermitian operator like $\mathbf{V}$ correspond to a physical observable?

In principle, yes, but per Ballentine's 3.37, the relationship between $\mathbf{V}$ and $\mathbf{Q}$ that justifies the term "velocity operator" for $\mathbf{V}$ only applies to expectation values. And the expectation value of $\mathbf{V}$ can only be interpreted as a "speed" for a state which is, to a good approximation, classical. But the state of an electron (or muon, for that matter) in a bound hydrogen atom is not classical even in approximation. So you can't just assume that the expectation value of $\mathbf{V}$, which is what appears in the quantum virial theorem, has a valid physical interpretation as the "speed" of the electron (or muon). Particularly since, as I've already pointed out to the OP, you can't use non-relativistic QM to analyze a situation where relativity is significant, and $\mathbf{V}$ is a non-relativistic operator (and the Hamiltonian you wrote down is also non-relativistic).