Not too long ago, someone began to discuss the thinking and motivation behind the Lagrangian and its formalism for the Newtonian framework and an intuitive understanding of such formalism. Somehow, it ended in the case that the Lagrangian can be understood in terms of conservation of information (largely) without having to rely on other laws.

Is this line of argument correct? And, more importantly, would this mean that unitarity is a stronger statement than conservation of energy?


3 Answers 3


I apologsise for this being only a partial answer. I can answer the questions in your title and in the final paragraph, but the first paragraph seems like quite a different question, and for the moment I'm not sure of the answer. (Some more details would help, e.g. a link to the discussion you mention.)

To answer the question in the title, consider unitary evolution expressed in its most general form: $$ |\psi\rangle(t) = U(t) |\psi_0\rangle. $$ For a time-independent system $U(t)$ must have the property that $U(t_2)U(t_1) = U(t_2+t_1)$. This implies that $U(t) = A^t$ for some unitary operator $A$. One may express $A$ as $e^{-iH}$ for some Hermitian operator $H$, giving us $$ |\psi\rangle(t) = e^{-iHt} |\psi_0\rangle, $$ or, differentiating, $$ \frac{d}{dt}|\psi\rangle = -iH|\psi\rangle, $$ which is the Schrödinger equation. If one considers $H$ to be an observable and defines the energy as its expectation, it follows that the energy is conserved. Thus we can conclude that unitarity, together with time-independence, implies the conservation of energy.

In your final paragraph you also ask about the converse of this, i.e. does the conservation of energy imply unitarity? This depends on how you define energy. Lots of dynamical systems have conservation laws without being unitary, and if you define energy as being the thing that's conserved in these cases then you'd have to conclude that unitarity is a stronger statement than conservation of energy. On the other hand, if you restrict the term "energy" to mean the expectation of the Hamiltonian, then you've already assumed the Schrödinger equation in making the definition, so unitarity follows trivially. In that case the conclusion is that energy conservation and unitarity are equivalent, with neither statement being stronger than the other.

  • $\begingroup$ Hm. All right, thank you. I'm still wondering about the case for the Hamiltonian---how does describing the time evolution imply conservation of energy (that is, of course, the Schrodinger equation)? $\endgroup$ Nov 13, 2013 at 4:27
  • $\begingroup$ I'm not sure I understand your comment. Didn't I already show that describing the time evolution (as unitary and time-independent) implies the Schrödinger equation? $\endgroup$
    – N. Virgo
    Nov 13, 2013 at 5:06
  • $\begingroup$ Sorry, I should have been more careful in my above statement, but yes--my mistake. $\endgroup$ Nov 15, 2013 at 0:41

No. Unitarity guarantees that whenever the Hamiltonian exists, energy is a real number, but it does not imply energy conservation.

Writing $$|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle\text{,}$$ we should have the composition law $$U(t+t',t_0) = U(t+t',t')U(t',t_0)\text{.}$$ If $U$ is translation-invariant, $U(t+t',t_0+t') = U(t,t_0)$, then it follows that $U(t+t',0)$ $=$ $U(t+t',t')U(t',0)$ $=$ $U(t,0)U(t',0)$, so we can take $U(t,0) = A^t$ and use the fact that the logarithm of a unitary operator is skew-hermitian whenever it exists: $A = e^{-iH/\hbar}$ for some constant, hermitian $H$.

If it's not time-translation-invariant, we can't do this. Just as with classical mechanics, energy conservation and time translation are linked. However, for short time-translations $\delta t$, $$U(t+\delta t,t_0) = U(t,t_0) + (\delta t)\dot{U}(t,t_0) + {\mathcal O}(\delta t^2)\text{,}$$ and unitarity requires that, to first order in $\delta t$, $$\begin{eqnarray*} 1 &=& \left[U(t,t_0) + (\delta t)\dot{U}(t,t_0)\right]\left[U(t,t_0) + (\delta t)\dot{U}(t,t_0)\right]^\dagger \\ &=&1 + (\delta t)\underbrace{\left[\dot{U}U^\dagger + U\dot{U}^\dagger\right]}_0\text{.} \end{eqnarray*}$$ Thus the operator $\dot{U}U^\dagger$ is skew-hermitian, so defining $H = i\hbar \dot{U}U^\dagger$, $H$ must be hermitian, and in particular $$\begin{eqnarray*} U(t+\delta t,t_0) &=& U(t,t_0)U^\dagger(t,t_0)U(t,t_0)\\ &=& \left[1+(\delta t)\frac{H}{i\hbar}\right]U(t,t_0)\text{.} \end{eqnarray*}$$ Re-arranging and taking the limit as $\delta t\to 0$, $$i\hbar\frac{\partial}{\partial t}U(t,t_0) = HU(t,t_0)\text{,}$$ which is the Schrödinger equation for the time-evolution operator. To usual Schrödinger equation follows immediately by multiplication by the state ket.


On the flip-side, energy conservation, in the sense of a time-independent Hamiltonian, doesn't guarantee unitarity. For a trivial example, take a nonrelativistic particle with Hamiltonian $\hat{\mathcal H} = \frac{1}{2m}{\hat{p}}^2 + V(x)$ with complex but time-independent potential. Then probability is not conserved, as (for the right imaginary sign) the particle has an exponentially decaying probability of being around.

  • $\begingroup$ Very complete answer. I'll probably be marking this as the solution, but I'm waiting until I get a response from Nathaniel. $\endgroup$ Nov 13, 2013 at 4:33

As showed in the answer of Stan Liou, the unitarity of the evolution operator $U$ does not imply conservation of energy, it only implies that the hamiltonian $H(t)$ is an hermitian operator and an observable, and so any measurable energy value is a real number.

More generally, the fact that energy and momentum operators are observable, come, "in fine", that the classical lagrangian, and the classical action, are real quantities.

The Unitarity of $U$ implies (for a pure state), the conservation of the information (along time variations), or, equivalently, the conservation of the total probability (along time variations).

On a other way, conservation of energy, and conservation of momentum, come respectively from a time-invariance , and a space-invariance of the lagrangian.


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