Unitary transformations are defined as transformations that preserve the norm of state vectors. In linear algebra, this is translated as preserving the inner product of the elements of the vector space in which the unitary transformation acts.

In optomechanical systems (one I am studying currently) and many other systems, we generally start with a standard Hamiltonian (with maybe an extra term or so arising from our system choice) and apply a unitary operator o the hamiltonian.

The new transformed hamiltonian beings out some new features of the system. We then go on to analyze the system such as calculating rate equations etc for this system using this new hamiltonian.

How do we go from knowing that this unitarily transformed hamiltonian preserves the norm to the fact that the rate equations will yield the same equations of motion and that the whole system can now be effectively studied with this new hamiltonian?

So, what does a unitary transformation in general preserve when it acts on operators and vectors, and in what cases and to what extent the system hamiltonian can be replaced by a unitarily transformed hamiltonian?

PS: Previously, this question didn't bother me because I just thought of unitary transformations as changes of basis (from one ordered orthonormal basis to another orthonormal basis) and thought ...well this is the same as looking at the system from a different co-ordinate system so to say and therefore nothing is changing. But I now realize that a more rigorous definition as opposed to my dumb laid back understanding is that these are transformations that preserve the norm.

I am looking for an answer that takes all the above doubts, misunderstandings and points into consideration.

  • 3
    $\begingroup$ Your "PS" appears entirely correct to me - a unitary transformation is just a change from one orthonormal basis into another. What more are you looking for here? $\endgroup$
    – ACuriousMind
    Commented Oct 22, 2021 at 4:41
  • $\begingroup$ Why is it then that the rigourous (formal) definition of unitary transformations only talks about them preserving the norm ? Do the rate equations etc automatically get preserved when we constrain the norm to get preserved? $\endgroup$
    – Lost
    Commented Oct 22, 2021 at 11:33
  • $\begingroup$ 1. There are different formal definitions of unitary operators, not all of them are about preserving the norm (though of course they all imply it since they are equivalent definitions). 2. An operator that preserves the norm also necessarily sends orthonormal bases to orthonormal bases.. $\endgroup$
    – ACuriousMind
    Commented Oct 22, 2021 at 12:41
  • $\begingroup$ About the PS : this is basically the difference between active and passive transformations/change of basis. $\endgroup$ Commented Oct 22, 2021 at 14:14
  • $\begingroup$ @ACuriousMind Okay got it. $\endgroup$
    – Lost
    Commented Oct 23, 2021 at 7:34

1 Answer 1


As I pointed out in the comments, this boils down to the distinction between active and passive transformations.

Active transformations : Consider a Hilbert space $\mathcal H$. A unitary operator $U : \mathcal H \to \mathcal H$ is an active transformation : it actually maps a ket $|\psi\rangle$ to another ket $U|\psi\rangle$.

Passive transformations : On the other hand if we take two orthonormal bases $\{ |n\rangle_1 : n\in\mathbb N\}$ and $\{ |n\rangle_2 : n \in \mathbb N\}$, they are related by a matrix $(U_{nx})$ so that : $$|n\rangle_2 = \sum_n U_{mn}|m\rangle_1$$ Then, any ket can be expanded as : $$|\psi\rangle = \sum_n \psi^1_n|n\rangle_1 = \sum_n \psi^2_n|n\rangle_2$$ where the coordinates in the different bases are related by : $$\psi^1_n = \sum_m U_{nm}\psi^2_m$$ The matrix is unitary in that : $$\sum_k U_{nk} (U_{mk})^* = \delta_{nm} \qquad\text{ and }\qquad \sum_k (U_{kn})^*U_{km} =\delta_{nm}$$ This is a passive transformations : a change of basis, so that the coordinates are changed even though the kets themselves are not.

Equivalence between the two and their use in quantum mechanics

Given two orthonormal bases as before, there is a unique unitary operator $U:\mathcal H\to \mathcal H$ such that $U|n\rangle_1 = |n\rangle_2$ for every $n$.

Then, for any operator $A$, we have : \begin{align} {}_2\langle n|A|m\rangle_2 &= {}_1\langle n|U^\dagger AU|m\rangle_1 \end{align} That is : the matrix elements of $A$ in the basis $2$ are the same as that of $U^\dagger A U$ in the basis $1$.

Physical quantities and law do not depend on the basis we use, so the passive transformation shouldn't change the physics. However, doing a change of basis in QM is quite cumbersome (as the Dirac braket notation is not particularly well suited for this).

Sometimes however, the basis where the problem is naturally stated (say basis $1$) is not the basis where computations are easily done (say basis $2$). This means that the matrix elements of some operators are nicer in basis $2$. Instead of doing the (physical) passive transformation, we can do an active transformation : changing kets but keeping the same basis $1$.

Edit : Consider a $1D$ spinless particle with Hamiltonian $H = \frac{p^2}{2m} + V(x)$. The active transformation corresponding to a translation by $a$ is implemented by the unitary operator $U$ such that : $$U|x\rangle = |x+a\rangle$$ We can then check that the following equations are true : \begin{align} U|p\rangle &= e^{ipa}|p\rangle \\ Ux U^\dagger &= x - a \\ UpU^\dagger &= p \\ UHU^\dagger &= \frac{p^2}{2m} + V(x-a) \end{align}

What this means is :

  • we can think of what we did as a "passive" transformation by defining new position, momentum , etc. operators by : $A' = UAU^\dagger$ ie $p' =p$, $x' = x-a$ so that $H' = \frac{p'^2}{2m} + V(x')$, and we effectively did nothing.
  • or think of it as an active transformation : we actually moved the system. So we keep the operators $x$ and $p$ and use the new hamiltonian $H= \frac{p^2}{2m} + V(x-a)$
  • $\begingroup$ "Physical quantities and law do not depend on the basis we use, so the passive transformation shouldn't change the physics." $\endgroup$
    – Lost
    Commented Oct 23, 2021 at 11:22
  • $\begingroup$ (Cont'd) Does an active transformation change the Physical quantity or physics of the system? $\endgroup$
    – Lost
    Commented Oct 23, 2021 at 11:23

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