I'm reading J. Schwinger's book "Quantum Kinematics and Dynamics" and I'm trying to make sense of his formulation of his famous quantum action principle. In essence, he starts from considering arbitrary variations of the the quantum amplitude $\langle a, t_1 |b, t_2 \rangle$, where $|a, t_1 \rangle$ is an eigenvector, corresponding to eigenvalue $a$, of some arbitrary but fixed Hermitian operator $A(t_1)$ (considered in the Heisenberg picture of QM), and $|b, t_2 \rangle$ is an eigenvector, corresponding to eigenvalue $b$, of another arbitrary but fixed Hermitian operator $B(t_2)$. An arbitrary variation of the amplitude reads, obviously, $$ \delta (\langle a, t_1 |b, t_2 \rangle)= (\delta\langle a, t_1 |)|b, t_2 \rangle + \langle a, t_1 |(\delta |b, t_2 \rangle) $$ where $\delta\langle a, t_1 |$ and $\delta |b, t_2 \rangle$ are arbitrary variations of the eigenbra and eigenket.

Now comes the puzzle. Schwinger writes $$\delta |b, t_2 \rangle = -iG_b(t_2)|b, t_2 \rangle$$$$\delta\langle a, t_1 | = iG_a(t_1)\langle a, t_1 |$$where he says $G_a(t_1)$ and $G_b(t_2)$ are infinitesimal Hermitian operators. I don't understand as to why an arbitrary infinitesimal variation $\delta |b, t_2 \rangle$ should be generated by a Hermitian operator, as if only unitary infinitesimal variations (instead of general ones) are considered, and not by a more general operator.

Could you, please, help clarify as to why the generators must be Hermitian for the most general variation $\delta |b, t_2 \rangle$?

  • $\begingroup$ Could you include how his variation $\delta(<a,t_1|b,t_2>)$ is defined(if he defines at all)? $\endgroup$
    – user110373
    Aug 5, 2016 at 3:30
  • $\begingroup$ No, he doesn't define it more specifically, other than saying that it is arbitrary. I have reproduced all that he wrote. $\endgroup$ Aug 5, 2016 at 3:38

2 Answers 2


It is natural to endow any time-evolution of isolated systems with the natural abelian group structure of reals, i.e. $E(t)E(s)=E(t+s)$, $E(t)^{-1}=E(-t)$. This is because we want our mathematical definition of evolution to behave accordingly like a reversible physical time evolution that can be applied step-by-step.

On the other hand, it is also natural to assume the quantum time-evolution to be a group of linear operators acting on the Hilbert space of the theory, and that at each time it preserves probabilities (because of our probabilistic physical interpretation of wavefunctions).

With these assumptions, it follows that the quantum evolution $(E(t))_{t\in\mathbb{R}}$ should be a group of unitary operators, i.e. satisfying:

  • $(\forall t\in\mathbb{R})$ $E(t)$ is a unitary operator on the Hilbert space $\mathscr{H}$ of wavefunctions;

  • $(\forall t,s\in\mathbb{R})$ $E(t)E(s)=E(t+s)$ (i.e. $E(t)^*=E(-t)$).

Finally, it is also desirable that such evolution is strongly continuous, i.e.

  • $(\forall s,t\in\mathbb{R})(\forall\psi\in\mathscr{H})$ $\lim_{s\to t}E(s)\psi=E(t)\psi$ (where the limit holds in the strong topology of $\mathscr{H}$);

because we want our probability distribution to behave smoothly as time passes.

Combining these three properties, we obtain a strongly continuous group of unitary operators, and by Stone's theorem its infinitesimal generator must be a self-adjoint operator.

The above reasoning can be extended to any more general symmetry transformation that behaves like time-evolution (i.e. has an underlying abelian group structure, preserves quantum probabilities and is smooth).

  • $\begingroup$ +1....I think you already know, but strongly continuity is equivalent to measurability of all the maps $r \to \langle \psi| E(r) \phi \rangle$ if the Hilbert space is separable. $\endgroup$ Aug 5, 2016 at 9:23
  • $\begingroup$ @ValterMoretti Thanks for the further insight ;-) I knew the result (due to von Neumann if I recall correctly); but I don't know of any of its applications. Do you know if it is actually used to prove the strong continuity property indirectly? $\endgroup$
    – yuggib
    Aug 5, 2016 at 9:38
  • $\begingroup$ @yuggib Thanks, but these things are well-known. If you read Schwinger's book, you'll find a completely original approach to QM and time evolution. Instead of postulating a time-evolution operator, he postulates the quantum action principle, from where he deduces the dynamical evolution equations. My question refers to things that are not postulated but seem evident to Schwinger. I would appreciate if you could answer my question for an arbitrary Hermitean operator, since a general variation $\delta |b, t_2\rangle$ is not due only to time evolution. $\endgroup$ Aug 5, 2016 at 9:43
  • $\begingroup$ It's clear that if one assumes delta is due only to time evolution it must be Hermitian, with it's integral unitary, as some have said above. All that that requires is that the probabilities of a certain state be preserved, ie, inner products of states with themselves. But of course delta is just an arbitrary variation, clearly G is not Hermitian if probabilities are not preserved. Seems to me you need some extra assumption, maybe Schwinger pre-assumes it. $\endgroup$
    – Bob Bee
    Aug 5, 2016 at 16:52

A short answer would be that because states are transformed via unitary transformations, infinitesimal transformations would be given by i times a hermitian operator.

For a longer answer one can use some model for the states where one expresses an arbitrary state $|a,t\rangle$ as a corresponding unitary operator $U(a,t)$ that converts some fixed reference state $|\Omega\rangle$ into that state

$|a,t\rangle=U(a,t) |\Omega\rangle . $

The variation of the state then becomes the variation of the unitary operator

$\delta|a,t\rangle \rightarrow \delta U(a,t) . $

Any unitary operator can be expressed in terms of an exponential function that contains hermitian matrices in its argument

$U(a,t)=\exp(i{\bf k}(a,t)\cdot{\bf A}) , $

where ${\bf k}(a,t)$ is a vector of real-valued parameters and ${\bf A}$ is a vector of hermitian matrices (think generators).

So the variation would then take the form

$\delta U(a,t)=i(\delta{\bf k}(a,t)\cdot{\bf A}) U(a,t) . $

Allow this to operate on the reference state and we end up with

$\delta|a,t\rangle = i G(a,t) |a,t\rangle , $


$G(a,t) = \delta{\bf k}(a,t)\cdot{\bf A} $

is hermitian.

  • $\begingroup$ Thank you very much for your answer! I still have one question before I vote it. From your derivation, it means that starting from some fiducial vector $|\Omega\rangle$ one can generate the whole set of eigenvectors $|a,t\rangle$ of any Hermitean operator $A(t)$ by applying a group of unitary operators to the same fiducial vector $|\Omega\rangle$. I cannot find a counterexample off the top of my head, but is it true for any Hermitean operator? This is my only concern. Thanks! $\endgroup$ Aug 5, 2016 at 6:02
  • $\begingroup$ I would probably not be able to give a rigorous proof for it, but I can give an argument. All the states belong to some space on which these Hermitean operators are defined. Hence, all the eigenvectors of these operators will lie in this space. Now I'll conjecture (don't know how to proof it off the top of my head) that for any pair of states in this space there will always exist a (often more than one) unitary operator that will map one of the states into the other. So it should work if one state is the fiducial vector and the other is an eigenvector of some Hermitean operator. $\endgroup$ Aug 5, 2016 at 9:37
  • $\begingroup$ The fiducial vector $|\Omega\rangle$ cannot be different for different eigenvectors $|a,t\rangle$, since if it depends on them, then it should be also varied when taking $\delta |a,t\rangle$, and we go nowhere. If $|\Omega\rangle$ is the same for all eigenvectors, then for me it's not at all obvious as to why these eigenvectors are related to $|\Omega\rangle$ by unitary transformations. For a general Hermitean operator, $A(t)$, this very strong statement has to be proven. $\endgroup$ Aug 5, 2016 at 10:10
  • $\begingroup$ Let $|a\rangle$ and $|b\rangle$ be 2 arbitrary vectors of norm 1 in a Hilbert space. Then, $|a\rangle = U_{ab}|b\rangle$, where $U_{ab} = |a\rangle\langle b|$ and $U_{ab}^{\dagger} = |b\rangle\langle a|$. Then, $U_{ab} U_{ab}^{\dagger} \neq U_{ab}^{\dagger}U_{ab} \neq \bf{I}$. $\endgroup$ Aug 5, 2016 at 11:53
  • 1
    $\begingroup$ I can think of a simple geometrical argument that would show that all pairs of vectors are related by a unitary. If I consider a vector space then two such vector defines a two dimensional subspace. (In general there would be an infinite number of two-dimensional subspaces that contain these two vectors, but let's restrict ourselves to the one produced by all the linear combinations of these two vectors). Then the required unitary is simply the rotation in this plane that would take you from one vector to the other. $\endgroup$ Aug 5, 2016 at 13:03

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