In the context of unitary transformations as a change of basis in Hilbert space, Gottfried and Yan wrote in their book on quantum mechanics that expressing a unitary operator $U$ in terms of a Hermitian operator $Q$ is useful in a multitude of situations, such as in perturbation theory and in the description of continuous symmetry. Examples of useful forms are $$U=e^{iQ},$$ $$U=\frac{1+iQ}{1-iQ}.$$ What are some specific cases where such forms are useful? What is the benefit of expressing something in terms of a Hermitian operator?

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    $\begingroup$ That first question you're asking is just looking for a list, I think that should be taken out. (Or rephrase to say you're just looking for an example or two, if that's what you want) $\endgroup$
    – David Z
    Jul 4, 2012 at 1:48

3 Answers 3


The first form is the main thing--- it tells you the relation between infinitesimal generators and finite transformations of the quantum state space. If you have a continuous family of unitary transformations U(t) such that U(0) is the identity, the infinitesimal form at small t is that

$$ U(\epsilon) = 1 + \epsilon \dot{U} $$

Where $\dot{U}={dU\over dt}$ is anti-Hermitian, since the condition of unitarity is that $U^\dagger=U$, which implies to first order in $\epsilon$ that $\dot{U}^\dagger + \dot{U}=0$.

The standard thing in quantum mechanics is to multiply $\dot{U}$ by i and define $H=i\dot{U}$ (or by minus i, depending on the transformation), so that H becomes Hermitian instead of anti-Hermitian, so that the eigenvalues are purely real instead of purely imaginary. This is to conform to the conventional definition of observables as matrices with real eigenvalues. This is purely a convention, since any matrix with a commuting real and imaginary part has orthogonal eigenvectors, and so could have been considered an observable quantity. But the current convention is a historical artifact of the early days of quantum mechanics--- Heisenberg wanted observables to by analogous to real valued functions on phase space, so he defined his observables to be Hermitian. There is little to gain by bucking this convention, since if you have a quantity you consider a complex observable, the real and imaginary parts are separately real observables.

If the generator is constant, the result is that

$$ {dU\over dt} = -iH$$

and this is solved by

$$ U(t) = e^{iHt}$$

This is the main relationship between Hermitian observables and unitary transformations of the state space. This shows up everywhere:

  • $ e^{-iHt}$ where $H$ is the Hamiltonian (for a system without time-dependent constraints, just the energy) is the time evolution operator. It's matrix elements between two position states is the propagation kernel.
  • $ e^{iPt}$ where $P$ is the momentum makes a translation on phase space.
  • $ e^{iQt}$ where Q is the charge operator makes a field rotation which implements global gauge transformations.

The relation between observables and transformations of phase space has a classical analog in the canonical transformations, which are made when you use a classical observable (a function on phase space) as a Hamiltonian, and use Hamilton's equations to move phase space forward in time.

The function that you use to do this is called the infinitesimal generator of the canonical transformation. The relation in classical mechanics is that for any function on phase space $f(p,q)$ you get a Hamiltonian flow

$$ \dot{q} = {\partial f \over \partial p}$$ $$ \dot{p} = - {\partial f \over \partial q}$$

Which is more abstractly called the flow of the vector field made by acting the symplectic form on the gradient of f. This produces a canonical transformation at any time t, since Hamilton's equations preserve the phase-space structure for any Hamiltonian, even an arbitrary one like f. The quantum analog of the relation between the infinitesimal and finite canonical transformation is the exponential thing you write down.

The other form

$$ U= {1 + i Q\over 1- iQ}$$

is nowhere near as useful.

  • $\begingroup$ Can you make clear your first part of explanation? That: Where U˙=dUdt is anti-Hermitian, since the condition of unitarity is that U†=U, which implies to first order in ϵ that U˙†+U˙=0 Unitary has inverse in the definition. U†=U^-1. I did not understand your explanation. Can you explai? Thanks. $\endgroup$
    – AlexandreH
    Mar 12, 2013 at 21:18
  • $\begingroup$ It's anti-Hermitian, but when a matrix is anti-hermitian, you multiply it by i and it becomes Hermitian, and this is what physicists traditionally do. If you multiply the Hermitian thing by -i you go back to the anti-Hermitian thing. This part is relatively straightforward, I think. But if it is still confusing, I'll change it. $\endgroup$
    – Ron Maimon
    Mar 13, 2013 at 1:07
  • $\begingroup$ Ron, Shouldn't the exponent in "this is solved by U(t)=e^(iHt)" be negative, as in the expression three lines further down? $\endgroup$
    – iSeeker
    Oct 5, 2019 at 10:43

Hermitian operators form a linear space over reals. That means if you add two given Hermitian operators (or multiply a given Hermitian operator with a real number) you again get a Hermitian operator. Unitary operators on the other hand do not have such linear structure (they rather have a group structure w.r.t. multiplication). It has been observed that linear space structure is usually (though may be not always) easier to work with as compared to group structure (addition is usually easier to handle than multiplication). So usually if one is given with some problem about unitary operators one likes to first translate it in terms of Hermitian operators. This practice is in fact more generally followed in most of group theory; for example instead of working with group of orthogonal matrices one would prefer to work with their linear counterpart which is space of real antisymmetric matrices. In quantum mechanics preference to work with Hermitian operators (instead of unitary operators) becomes even stronger because it is these operators which are interpreted to physical observables.


As mentioned by dushya, Hermitian matrices are much easier to handle since they form a linear space. Unitary operators are far more difficult to construct (and to use outside of purely theoretical considerations).

The exponential representation has many advantages, but is also somewhat difficult to use as to find the exponential you need to solve a operator-valued differential equations to go from $Q$ to $U$, and for the converse direction you need a log, which usually requires a full spectral resoluution.

The Cayley transform $U=(1+iQ)/(1-iQ)$ is easier to use as the inverse transform has nearly the same shape, and both transformations just require the solution of linear operator equations. It is used for example to unitarize matrices (go from an approximate unitary matrix to its Cayley transform, symmetrize, and transform back to get an exact unitary matrix.)

Both exponential transform and Cayley transform work generally in a neighborhood of the identity but may fail far away from it.

  • $\begingroup$ About the Cayley transform you mention; is there a pedagogically resource how to see the process of approx. unitary matrix -> Cayley transform - > symmetrization -> back transformation, applied to a physicak example? $\endgroup$ Aug 7, 2018 at 17:56
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    $\begingroup$ @user2820579: I don't know of a particular resource, it is usually applied in context. But it is easy to apply to scattering cross sections for 1DOF toy problems, where the S-matrix is diagonal and just consists of a phase. Simple approximations (e.g. from truncated perturbation theory) are not unitary. In this case the simple solution would be to take the polar decomposition and to drop the radial part, but one can also do the above recipe and get algebraically simpler results of comparable quality. $\endgroup$ Aug 8, 2018 at 17:14

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