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In this Question I want to give a derivation of Hamiltons Principle of Stationary action, and my question to the community would be, whether my argument is flawed. The System I want to look at is (for simplicity) a particle moving in 1 dimension, so what I can observe about this particle is its position, x. For a Quantum Mechanical Treatment, I assume that the System will be in superposition of many states, and therefore the way to go is:

Assumption: The observable $X$ is represented by an operator $\hat{X}$ acting on a suitable hilbert space over the complex numbers. A state of the system is a vector (whiches norm is $1$) in this hilbert space, and it's decomposition into eigenstates of the operator $\hat{X}$ will give the probabilities of measuring the eigenvalues of the operator.

Also an Assumption? I'm not sure about that: Since we want to propability to be conserved, we require time evolution of the Operator $\hat{X}$ to be unitary: $\dot{\hat{X}}=\frac{i}{\hbar}[\hat{H}(t), \hat{X}]$, with a yet to determine operator $\hat{H}$. I'm not sure about wether this is an assumption or not, because the overall probability has to be conserved. Since I hereby state that states are stationary in time, while operators experience time evolution, I am in the Heisenberg picture, operators move in time, states do not.

Definition of another observable (which will later on turn out to have similar properties of what we usually call "momentum"): Given the observable $\hat{X}(t)$, we define $\hat{F}$ as the generator of translations of $X$, which means it should hold at every time that $[\hat{X},\hat{F}] = i\hbar $ and $\dot{\hat{F}}=\frac{i}{\hbar}[\hat{H}(t), \hat{F}]$.

By that definition, $F$ will generator infinitesimal c-number variations of $X$: $\hat{X}'(t) = \hat{X}(t) + \delta X(t) = \hat{X}(t) + \frac{i}{\hbar}[\hat{F}(t)\delta X(t), \hat{X}(t)]$. At the same time, $-\hat{X}(t)$ will be the generator of infinitesimal c-number variations of $\hat{F}$. We can write $\hat{H}(t) = \hat{\tilde{H}}(\hat{X}, \hat{F}, t)$ (without giving an explicit formula yet, the dependence could be no dependence at all).

Now I assume a variation of the Quantity $\dot{\hat{X}} \hat{F} - \hat{\tilde{H}}(\hat{X}, \hat{F})$. By that I mean $$\delta L = \dot{(\hat{X} + \delta X)} (\hat{F}+\delta F) - \hat{\tilde{H}}(\hat{X}+\delta X, \hat{F}+ \delta F) - (\dot{\hat{X}} \hat{F} - \hat{\tilde{H}}(\hat{X}, \hat{F}))$$ Doing some calculations, and using $\hat{\tilde{H}}(\hat{X}+\delta X, \hat{F})- \hat{\tilde{H}}(\hat{X}, \hat{F})= -\delta X \dot{\hat{F}}$ (and the same for $F$), we arrive at: $$ \delta \hat{L}(t) = \dot{\delta X} \hat{F} + \delta X \dot{\hat{F}} = \frac{d}{dt} ( \delta X \hat{F} ) $$

Choosing $\delta X(t_1) = \delta X(t_2) = 0$, we arrive at:

$$ \int_{t_1}^{t_2} \delta L(t) = \delta X(t_2) \hat{F}(t_2) - \delta X (t_1) \hat{P}(t_1) = 0 $$ Or written in its complete form: $$ \delta \int_{t_1}^{t_2} \dot{\hat{X}} \hat{F} - \hat{\tilde{H}}(\hat{X}, \hat{F}) = 0 $$ Where $\delta$ means variation of the operators $\hat{X}$ and $\hat{F}$ by c-number multiples of $\mathbb{1}$, and the variation of $\hat{X}$ is supposed to be 0 at the times $t_1$ and $t_2$. Which es exactly the principle of stationary action with a yet unkown quantity $F$ inside.

Then using $\frac{\partial \hat{\tilde{H}}}{\partial \hat{F}} = \frac{i}{\hbar}[\hat{\tilde{H}}, \hat{X}] = \dot{\hat{X}}$, one can perform a Legendre Transformation $F \rightarrow \dot{X}$, and arrives at the same principle, but formulated for a lagrange function that depends on $X$, $\dot{X}$, and (possible) higher derivations.
Is any of these arguments invalid? Did I make any further assumptions as the ones listed here?

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To me, the flaw lies in the very definition of your lagrangian $\hat{L}=\dot{\hat{X}}\hat{P}-\hat{\tilde{H}}$.

According to your definition, $\hat{L}$ is an operator. So what exactly do you mean by minimising it ? [For this purpose you would have to introduce one or several quantum states, and minimise e.g. the expectation value.] Another way to see this flaw is to ask what you mean by e.g. $\dot{\hat{X}}$ ? In the Schrödinger picture, $\hat{X}$ (and $\hat{P}$) are time independent operators, so the time derivative would be zero. Again, the dynamics (and hence the time derivative) only comes in with the quantum states $\psi(x,t)$.

The standard way to connect quantum mechanics to the least action principle is via Feynman's path integral. Its derivation indeed requires unitary time evolution plus the canonical commutation relations of $\hat{X}$ and $\hat{P}$, the difference being that you calculate the probability amplitude for the transition of a specific initial state $|i>$ to a specific final state $|f>$. You end up with the desired action integral, however in a wave-like exponent, i.e. $\exp[\frac{i}{\hbar} S]$. And this is where the essence of quantum mechanics lies: you don't forbid the system to explore paths away from least action, but the wave-like nature of $\exp[\frac{i}{\hbar} S]$ suppresses those paths by destructive interference.

As an example, think of the standard problem of a particle in a finite rectangular 1d well: Suppose the particle has energy $E$ smaller than the depth of the well $V_0$. Then, classically the particle couldn't get out, so its trajectory (given by the least action principle) should be the same regardless the depth of the well (as long as $E<V_0$ of course). Quantum mechanically, we know, however, that the depth of the well does affect the energy eigenstates. There must thus be some way for the quantum particle to explore the classically forbidden zone outside the well, and this is encoded in the states, not in the operators.

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  • $\begingroup$ The Action principle requires $\dot{\hat{X}} \hat{P} - \hat{H}$ to be stationary with respect to variations of $\hat{X}$ and $\hat{P}$. In that form, this can also hold for operators. $\dot{\hat{X}}$ . And I mean the time derivative of the operator, since I'm in the Heisenberg picture here. I'll make edits to my previous post, to point this out more clear. $\endgroup$ – Quantumwhisp Oct 30 '17 at 10:26
  • $\begingroup$ Sorry, I haven't seen your comment since I was editing my answer. Again, even in the Heisenberg picture, you will not get away with operators alone. Think of my example of the quantum well: in the Heisenberg picture, the time dependence goes to the operators, but you will need the states [which are just the solution of what is usually called the time independent Schrödinger eq.] Otherwise, all you get is some Bohr Sommerfeld like formalism. $\endgroup$ – Stesh Oct 30 '17 at 10:58
  • $\begingroup$ I think I spotted sth else: You state that "doing some calculations and using $\hat{\tilde{H}}(\hat{X}+\delta{X},\hat{P})-\hat{\tilde{H}}(\hat{X},\hat{P})=-\delta{X}\hat{\dot{P}}$...". How do you derive this without using the classical Hamilton eq $\frac{\partial H}{\partial X}=-\dot{P}$ ? Remember that Hamilton's eqs are derived from the Euler-Lagrange eqs which follow from the least action principle, so using them here would render your argument circular. $\endgroup$ – Stesh Oct 30 '17 at 11:52
  • $\begingroup$ What do you mean by "I will not get away with operators alone"? I'm not getting the last part of your answer. And I know about the Feynman Path Integral Formalism, but that's not what my question is about. In The Heisenberg Picture, the operators do obey the principle of stationary action as well (as operators!), and I'd like to derive this from the assumptions I have made. $\endgroup$ – Quantumwhisp Oct 30 '17 at 13:21
  • $\begingroup$ To your Last concern: I defined $\hat{P}(t)$ the way it satisfies this equation. (I defined it as the Operator for which the canonical commutation relation always holds, I can easily define it as well as the function that obeys this equation). We can now discuss wether this is a definition or a assumption, but I personally would say its a definition, because by defining it I don't make any further assumptions on the time evolution of $\hat{X}$. Even if I assume this, it wouldn't be a circular argument. It is wrong to say that the hamilton equations are based on the lagrangian formulation. $\endgroup$ – Quantumwhisp Oct 30 '17 at 13:32

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