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Is it possible to make a prediction for the eigenvalues of the Lagrangian-/action-operator of a quantum system if I know the ones of the Hamiltonian, the position- and momentum-operator?

Can I use the formula

$$ L = \sum_a \frac{1}{2} ( \dot{q}_a \cdot p_a + p_a\cdot \dot{q}_a )- H $$

to find the eigenvalues since the operators $H, p, \dot{q}$ do not commute with each other (and so they are not diagonizable with the same basis)? So the eigenvalues of $L$ are not $T-V$. Which orthonormal basis can we choose for the calculation instead?

I know that the Hamilton and Heisenberg equations also hold for this operator version (see Quantum kinematics and dynamics by J. Schwinger). So I can use

$$ \dot{q}=\frac{\partial H}{\partial p}$$

or a bit more general in terms of the Heisenberg equations:

$$ \dot{q(t)} = {\frac {i}{\hbar }}[H,q(t)]+\left({\frac {\partial q}{\partial t}}\right)_{H}$$

to reformulate the Lagrangian which then depends only on $q,p$ since $H(q,p)$:

$$ L(q(t),p(t)) = \sum_a \frac{1}{2} [{\frac {i}{\hbar }}[H,q_a(t)] \cdot p_a(t) + p_a(t) \cdot {\frac {i}{\hbar }}[H,q_a(t)]+ ]- H(q(t),p(t)) $$

But I also didn't know (with this Lagrangian) which eigenvectors I have to choose.

Since I'm studying Schwinger's quantum action principle I'm interestend in this eigenvalues because they should be minimized by the condition $\delta \hat{S} = 0$


marked as duplicate by Cosmas Zachos, heather, Jon Custer, Gert, JamalS Dec 30 '16 at 12:06

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  • $\begingroup$ Up to now nothing useful. Cause I can't find an orthonormal basis in which $p, H$ are diagonal if $H$ depends on $q$. Am I wrong? $\endgroup$ – Alpha001 Dec 27 '16 at 0:29
  • $\begingroup$ The thread you mentioned is not useful to me because I'm really interested in the eigenvalues of $L$ or $S$ (according to Schwinger's quantum action principle). $\endgroup$ – Alpha001 Dec 27 '16 at 0:32
  • $\begingroup$ You know how to write L in terms of (Heisenberg) operatrors, explicitly. Do it in your question. $\endgroup$ – Cosmas Zachos Dec 27 '16 at 0:45
  • $\begingroup$ Added some new informations. What exactly do you mean with "rewriting $L$ in terms of (Heisenberg) operators explicitly"? $\endgroup$ – Alpha001 Dec 27 '16 at 14:30
  • $\begingroup$ Then I only know that I can diagonalize the operator $L$ but not what the eigenvalues are in general (like some combination of the eigenvalues of $p,q,H$)? $\endgroup$ – Alpha001 Dec 27 '16 at 15:02

I will recast your problem, but, full disclosure, I will not attempt to answer open-ended unstated questions on Schwinger’s action principle, also see Milton 2005, or the actual mainstream equivalent formulations of Dirac’s seminal 1933 path integral paper which Feynman streamlined so magnificently that everyone in the identical question is admonishing you to not even ask.

I will narrowly recast your question to a more systematically answerable one. In the Heisenberg representation, ignoring red-herring explicit time dependent Schr operators for simplicity, $$ A(t)=e^{itH/\hbar} A e^{-itH/\hbar}\equiv U A U^\dagger. $$ Applying this to q(t) and p(t), and resolving the commutators for a conventional Hamiltonian $p^2/2 + V(q)$ complete your return calculation to the Lagrangian, $$ L= U (p^2/2 -V(q))U^\dagger ~. $$ Now, the key point that everybody is reminding everybody is that, in general, that parenthesis, and so L , does not commute with H, so it is not time-invariant, unlike H. So you may not commute the U on the left to annihilate its unitary inverse on the right, easily. The expression, in general, is a time-dependent mess that most would sensibly not deign to look at.

But, if you must, look at its spectrum at t =0, just the parenthesis. The (real!) eigenvalue spectrum of the expression in the Hermitian operator of the parenthesis, barring freak complications, is gotten by finding the spectrum of a bizarre inverted potential in any picture you like, e.g. the Schroedinger picture.

As a formal wisecrack, imagine your original Hamiltonian potential is the inverted oscillator Yuce, Kilic & Coruh Phys Scr 74 (2006) 114, also available on the arXiV , that is $V=-\omega^2 q^2$, so, then your L will be identical to the plain oscillator H of sophomore physics!

(Forget about your awful spectrum of your H, addressed in the refs provided.) Conclusion: the spectrum of this L is that of the plain oscillator, $\hbar \omega (n+1/2)$, being totally schematic and unashamedly cavalier about boundary conditions.

This is all to dramatize that, the spectral problem of L, in principle, is a freak Schroedinger problem that does not, and should not!, get anybody anyplace. (Which is the reason most people recoil from Schwinger’s variational principle and map it to Dirac-Feynman.)

But, to dispel an apparent technical misconception of yours, the (real) spectrum of a Hermitian operator need not reflect simultaneous eigenstates of the component operators involved in the operator. (So, in Sophomore physics, the eigenstates of the oscillator hamiltonian are neither eigenstates of p nor q, of course.)

  • $\begingroup$ Thanks for clearifying the question. It was clear to me that the spectrum of an operator is in general not the same as a combination of the underlaying operator-eigenvalues (the question was more like: since this with the orthonormal basis for all underlaying operators doesn't work is there any other possibility to calculate the spectrum simple with the information of the quantum system. But I agree my question was bad formulated. I had some hope that I can calculate the spectrum of $L=T-V$ in an easy way but as you wrote this is in general not possible. $\endgroup$ – Alpha001 Dec 28 '16 at 9:46

There is no operator corresponding to $\dot{q}$ so that the question seems meaningless to me.

Or better, what is the meaning of the operator $\dot{q}$ in terms of operators $p$ and $q$?

In classical mechanics, when dealing with the variational Hamiltonian principle, that $\dot{q}$ is the derivative of $q$ along a generic curve and not the the effective motion solution of Hamilton equations, which is found stationarizing the action. Passing to the quantum formulation there is no generic curve of operators $q=q(t)$ where the derivative can be computed. It cannot be the Heisenberg evolution of $q$, since we do not know that evolution at this stage of computations.

Perhaps the question may make sense, but not in this form. Everything should be written using only operators $p$ and $q$.

  • $\begingroup$ But the Hamilton equation still hold (Schwinger showed thi in his textbook: Quantum kinematics and dynamics). So we can set $\dot{q} = \frac{\partial H}{\partial p}$. By doing so we expressed $\dot{q}$ in terms of $q,p$. $\endgroup$ – Alpha001 Dec 27 '16 at 10:03

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