A quantum superposition is a wave function such as :

$$ |\psi\rangle = \alpha |\psi_1\rangle + \beta |\psi_2 \rangle $$

Unitary evolution will transfer these states to $|\psi'\rangle \to U|\psi\rangle$

My question is can unitary evolution transform a wave function psi to a 'greater' superposition of states, for instance

$$ \alpha |\psi_1\rangle + \beta |\psi_2 \rangle+ \kappa |\psi_3\rangle=U(\alpha |\psi_1\rangle + \beta |\psi_2 \rangle) $$

If this is not possible, then how does one create (in real life) a superposition of wave-functions if no unitary operator exists to create it from a simpler wave function ?

  • 5
    $\begingroup$ A 45 degree rotation sends $\hat{x} \mapsto \hat{x} + \hat{y}$ $\endgroup$
    – jacob1729
    Jan 29, 2021 at 14:19
  • $\begingroup$ If unitary time evolution couldn't do that then after measurement, given a system in a state $|\phi\rangle$, you would never be able to evolve in such a way that subsequent measurement produced a different state. Since the system would remain in $|\phi\rangle$ indefinitely and the inner product with any other eigenstate of that operator would be zero. $\endgroup$
    – Charlie
    Jan 29, 2021 at 14:23
  • $\begingroup$ When the price of bananas changes, can the new price be a sum of two other prices, or does the price have to be 'created' as a sum before it changes? $\endgroup$
    – WillO
    Jan 29, 2021 at 14:39

3 Answers 3


Well, I see no reason why the action of some arbitrary unitary operator should keep a vector in some subspace. A trivial way to see this is to realise that rotations in two dimensions are unitary, and you can certainly rotate the $\hat{\mathbf{x}}$ vector to some combination of $\mathbf{\hat{x}}$ and $\mathbf{\hat{y}}$ using a rotation!

It's very easy to construct an equivalent quantum example of this. A "physical" example of this is neutrino oscillations: neutrinos are created (and detected) as one of their flavour eigenstates ("electron", "muon", or "tau" neutrinos), but these are not eigenstates of the system's Hamiltonian (which are the so-called "mass" eigenstates). Thus, if an "electron" neutrino is created in the sun and travels to the Earth, it evolves unitarily into a superposition of the other states and so you could well say that: $$\hat{U}(t) |\psi_e\rangle = \alpha(t)\, |\psi_e\rangle + \beta(t) \,|\psi_\mu\rangle + \gamma(t) \,|\psi_\tau\rangle.$$

So even though the system was created as a pure "electron" neutrino, since this state was not an eigenstate of the Hamiltonian, it evolved into a superposition of other orthogonal states.

As to the second part of your question, I don't know how such states are created experimentally, but unitary operations are just one of many types of allowed linear operations, but not all quantum operations need to be unitary: the most notable example being measurements, which cause the wavefunction to collapse to some eigenstate of the observable being measured, and which are certainly not unitary! Only symmetry operations need to be unitary (or anti-unitary, see Wigner's theorem) in order to conserve probability.


Vector spaces don't have some elements that are superpositions and others that are not superpositions. Such a distinction depends on a choice of basis, which is a choice, not something built in to the vector space.

So your question would need to be changed somehow in order to make it a question that makes sense. One way to do so would be to change it to a question about density matrices.


Part of this question is what you consider to be a superposition of states. The word "state" in that phrase is ambiguous. It usually refers to the eigenstates of some observable that we want to measure, while the state of the system can be any linear combination of these. If we were interested in a different observable, that also means a different set of eigenstates and therefore a different kind of linear combination of them. In essence, a system state may be an eigenstate with respect to one observable, making it not a superposition in this eigenbasis, while the same state can be considered as a superposition when changing the observable and with it the eigenbasis.

As a simple example, in Schrödinger QM, the evolution operator is the exponential map of the Hamiltonian, i.e. energy operator, $U(t) = e^{-iHt/\hbar}$. If your initial state is an eigenstate of the Hamiltonian to an eigenvalue $\hbar\omega$, then time evolution will just add phase factors, $$|\psi(t)\rangle = e^{-iHt/\hbar} |\psi(0)\rangle = e^{-i\omega t} |\psi(0)\rangle.$$ If you measure the energy of this, you wouldn't consider the state to be in a superposition. But if you measure the position, your state is not a position eigenstate and you could describe it as a superposition of position eigenstates. After that measurement, you are in a position eigenstate, which is not an energy eigenstate, and unitary evolution will do more to this state than just add a phase factor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.