What does the word action mean in quantum physics?

[cronopio]

What does the word "action" mean in quantum physics?

Hi everybody,

I was reading about the application range of quantum mechanics. I found the effects of quantum mechanics are negligible where the action is on the order of the Planck constant. But, What does the word "action" mean?

Thanks

 

[atyy]

In "old quantum theory", the "action" was quantized. This was before Schroedingers equations was discovered, so it's been supplanted. However, it still provides good heuristics.

In classical mechanics, there are Hamiltonian systems that are "integrable", that is they can be written in terms of periodic motions. Technically, that means one can use "action-angle" vavriables. http://en.wikipedia.org/wiki/Action-angle_variables

In "old quantum theory", this "action" was postulated to be quantized. http://en.wikipedia.org/wiki/Old_quantum_theory

Nowadays, quantization usually depends on the commutation relations between canonically conjugate variables. "Canonical" mean "Hamiltonian", so the underyling formalism is still the same, but no longer restricted to integrable Hamiltonian systems.

However, the quantization of action is still useful in a field called "quantum chaos". See, for example, the chapter on semiclassical quantization in Cvitanović and colleagues's http://chaosbook.org/.

 

[cronopio]

In "old quantum theory", the "action" was quantized. This was before Schroedingers equations was discovered, so it's been supplanted. However, it still provides good heuristics.

In classical mechanics, there are Hamiltonian systems that are "integrable", that is they can be written in terms of periodic motions. Technically, that means one can use "action-angle" vavriables. http://en.wikipedia.org/wiki/Action-angle_variables

In "old quantum theory", this "action" was postulated to be quantized. http://en.wikipedia.org/wiki/Old_quantum_theory

Nowadays, quantization usually depends on the commutation relations between canonically conjugate variables. "Canonical" mean "Hamiltonian", so the underyling formalism is still the same, but no longer restricted to integrable Hamiltonian systems.

However, the quantization of action is still useful in a field called "quantum chaos". See, for example, the chapter on semiclassical quantization in Cvitanović and colleagues's http://chaosbook.org/.

Thank you. But I wasn't reading neither "old quantum theory" nor the "quantum chaos". I was reading about "quantum mechanics non-relativistic". I think the action is concerned with angular momentum. However, I'm not sure.

 

[atyy]

Thank you. But I wasn't reading neither "old quantum theory" nor the "quantum chaos". I was reading about "quantum mechanics non-relativistic". I think the action is concerned with angular momentum. However, I'm not sure.

Could you describe what you were reading in more detail?

 

[cronopio]

Could you describe what you were reading in more detail?

Yes, of course:

"...Action has the same dimensions as Planck's constant, h. When the action is smaller or of similar size to h, quantum mechanics rules. When the action is much larger than h, quantum democracy gives way to the more rigid phenomena that we term classical mechanics. Thus classical mechanics emerges from the fundamental quantum mechanics when the action is large compared to Planck's h..."

 

[atyy]

Yes, of course:

"...Action has the same dimensions as Planck's constant, h. When the action is smaller or of similar size to h, quantum mechanics rules. When the action is much larger than h, quantum democracy gives way to the more rigid phenomena that we term classical mechanics. Thus classical mechanics emerges from the fundamental quantum mechanics when the action is large compared to Planck's h..."

Yes, that's the same "action" as in old quantum theory, which is the "action" in "action-angle" variables for integrable Hamiltonian systems. The "action" is the integral of position multiplied by momentum, so it has the same units as angular momentum, which is why sometimes it is conceived as "angular momentum" of some sort.

The rule you quote that when the action is large, things become classical, is a rule of thumb, as is "old quantum theory". The idea underlying the rule of thumb is the Bohr atom, which is a planetary model of the atom, with quantized angular momentum or "action". As one goes to larger and larger orbits, the discrete steps of angular momentum should become small compared to the angular momentum of the orbit, so things will appear classical.

It's only a rule of thumb, since the Bohr quantization gave way to Schroedingers equation (which miraculously reproduced the Bohr results in the special case of hydrogen).

 

[cronopio]

Yes, that's the same "action" as in old quantum theory, which is the "action" in "action-angle" variables for integrable Hamiltonian systems. The "action" is the integral of position multiplied by momentum, so it has the same units as angular momentum, which is why sometimes it is conceived as "angular momentum" of some sort.

That's a good answer for me. Finally, what is difference betweeen that action and this:

Action= ∫Ldt


where L is the lagrangian. Are there any differences?

 

[atyy]

That's a good answer for me. Finally, what is difference betweeen that action and this:

Action= ∫Ldtwhere L is the lagrangian. Are there any differences?

I'm not aware of a straightforward link between these two "actions".

To me the link is indirect. Let me keep to classical mechanics. The action with the Lagrangian is used in the Lagrangian formulation of mechanics. From the Lagrangian, one can obtain the Hamiltonian. In the special case of an integrable ("integrable" is a technical tern about the number of conserved quantities) Hamiltonian system, the system becomes essentially periodic in "action-angle" variables. So I think of these two actions as basically unrelated.

 

[vanhees71]

The most straightforward analysis of the meaning of the action in quantum theory is the path-integral approach. If you want to calculate the transition amplitude
$$U(t,x;t',x')=\langle{x}|\exp(-\mathrm{i} \hat{H} t)|x' \rangle \qquad (1)$$
by using
$$\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})$$
and the approximation
$$\exp(-\mathrm{i} \hat{H} \delta t)=\exp[-\mathrm{i} \hat{p}^2/(2m)] \exp[-\mathrm{i} V[\hat(x)]+O(\delta t^2)$$
by inserting many infinitesimal time evolution kernels into (1) you get the path integral
$$U(t,x;t',x')=\int_{(t,x;t',x')} \mathrm{D} x \int \mathrm{D} p \exp \left \{ \int_{t'}^t \mathrm{d}t \; \mathrm{i} [\dot{x} p-H(x,p)] \right \}.$$
The integral over the spatial trajectories is over all paths connecting $x'$ with $x$ at times [itex]t'[/tex] and $t$, while the momentum integral is unrestricted. <br /> <br /> Now you can integrate out the momentum-path integral, because it is a Gaussian integral, and you end up with<br /> $$U(t,x;t'x')=\int_{(t,x;t',x')} \mathrm{D} x \exp(\mathrm{i} S[x]),$$<br /> where<br /> $$S[x]=\int_{t'}^{t} \mathrm{d} t \left[\frac{m}{2} \dot{x}^2-V(x) \right ].$$<br /> If now $S \gg 1$ (in normal units meaning $S \gg \hbar$) you can use the stationary-phase approximation to approximately evalute the path integral. This shows that in this case the main contribution comes from the stationary point of the action functional, which is the classical trajectory of the particle.[/itex]

 

[cronopio]

and the approximation
$$\exp(-\mathrm{i} \hat{H} \delta t)=\exp[-\mathrm{i} \hat{p}^2/(2m)] \exp[-\mathrm{i} V[\hat(x)]+O(\delta t^2)$$

How can you do it?

 

[vanhees71]

Just take the expansion of the exponential on both sides up to linear orders of $\delta t$. Of course it's not that simple for higher orders, because the kintic-energy operator doesn't commute with the potential-energy operator.

 

[cronopio]

$$\exp(-\mathrm{i} \hat{H} \delta t)=\exp[-\mathrm{i} \hat{p}^2/(2m)] \exp[-\mathrm{i} V[\hat(x)]+O(\delta t^2)$$

Just take the expansion of the exponential on both sides up to linear orders of $\delta t$. Of course it's not that simple for higher orders, because the kintic-energy operator doesn't commute with the potential-energy operator.

$$\hat{1} + (-\mathrm{i} \hat{H} \delta t) + O(\delta t^2)=\hat{1} + [-\mathrm{i} \hat{p}^2/(2m) -\mathrm{i} V[\hat(x)]+O(\delta t^2)$$

I don't understand it. Why are there an equal?

 

[atyy]

Just a quick note.

The action that vanhees71 mentioned in post #9 is the quantum version of the classical action mentioned by cronopio #7. Both are associated with the Lagrangian formulation. As vanhees71 noted, the classical "minimize the action" principle is recovered as the "stationary phase" or "saddlepoint approximation" of the quantum path integral. The "saddlepoint approximation" is the first term in an expansion, with higher terms usually multiplied by ħ. So if ħ is small, things are very classical. So the series expansion of the action of the Lagrangian (path integral) formulation gives a quantitative way of saying how good the classical approximation is, which was vanhees71's point.

As far as I know, the action in the Lagrangian formulation is not directly related to the action in old quantum theory, which is associated with the Hamiltonian formulation. The association is indirect, via the relationship between the Lagrangian and Hamiltonian formulations classically. In quantum mechanics, the relationship between Lagrangian (path-integral) and Hamiltonian (canonical commutators) formulations requires a condition called Osterwalder-Schrader positivity.

 

[vanhees71]

$$\hat{1} + (-\mathrm{i} \hat{H} \delta t) + O(\delta t^2)=\hat{1} + [-\mathrm{i} \hat{p}^2/(2m) -\mathrm{i} V[\hat(x)]+O(\delta t^2)$$

I don't understand it. Why are there an equal?

Argh :-(. I forgot the $\delta t$'s on the right-hand side. Sorry. The correct equation is
$$\exp[-\mathrm{i} \hat{H} \delta t]=\exp[-\mathrm{i} \hat{p}^2 \delta t/(2m)] \exp[-\mathrm{i} V(\hat{x}) \delta t]+\mathcal{O}(\delta t^2).$$
Now, because of $\hat{H}=\hat{p}^2/(2m)+V(\hat{x})$ you see that indeed in linear order in $\delta t$ both sides of the equation are equal.

 

[vanhees71]

Just a quick note.

As far as I know, the action in the Lagrangian formulation is not directly related to the action in old quantum theory, which is associated with the Hamiltonian formulation. The association is indirect, via the relationship between the Lagrangian and Hamiltonian formulations classically. In quantum mechanics, the relationship between Lagrangian (path-integral) and Hamiltonian (canonical commutators) formulations requires a condition called Osterwalder-Schrader positivity.

I'm not so familiar with these mathematical details. However, it is very important to keep in mind that quantum theory is based on the Hamilton formalism. Only if the action is quadaratic in the canonical momenta the Lagrangian path integral leads to correct results. Otherwise, at best you may evaluate another model than you intend to calculate when using the naive Lagrangian path integral or you even violate the unitarity of time evolution. So you always must carefully start from the Hamiltonian formulation.

The action-angle variables are of course just special choices of canonical coordinates, which made the Bohr-Sommerfeld quantization possible. This shows already that this theory is applicable to the rare cases of integrable systems, where can find a complete set of action-angle variables to parametrize phase space. Full quantum theory is of course not limited in such a way.