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I was reading Quantum Computation explained to my mother, which makes the following claim

Postulate 2 A closed physical system in state V will evolve into a new state W , after a certain period of time, according to W = UV where U is a n × n unit matrix of complex numbers.

Here, V is a column matrix with n rows. Can anybody justify this assumption?

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2 Answers 2

up vote 3 down vote accepted

Dear Casebash, the statement above is a "postulate of quantum mechanics" (note that it is even called in this way) - a fundamental "axiom" or "assumption" of any theory that wants to be called "a quantum mechanical theory". There are about 5-10 general postulates of quantum mechanics and the logic of quantum mechanics only starts to make sense to someone once he understands all of them. So it is counterproductive to take individual postulates out of the context.

This particular postulate, referred to as the "linearity of the wave functions", guarantees that the wave functions can only be constrained by linear equations. The set of allowed values of the wave functions is known as the "Hilbert space" and the postulate above guarantees that the set (the Hilbert space) must be a linear vector space and all physically meaningful operations, including the values of observables and evolution, have to be given by linear operators on this Hilbert space.

For infinitesimal evolution in time, the evolution of a wave function or, more generally, a "state vector" is dictated by Schrödinger's equation, $$i\hbar \frac{d}{dt} |\psi\rangle = H | \psi\rangle.$$ If the linear equation above is right, one can also mathematically prove that $$|\psi(t_1)\rangle = U | \psi(t_0)\rangle.$$ In fact, we may exactly calculate what $U$ is in terms of the Hamiltonian $H$: it is $$ U = \exp(H(t_1-t_0) / i \hbar) $$ where we have to exponentiate an operator - which is possible. More generally, if the norm $\langle \psi| \psi\rangle$ is conserved, and it should be because it may be interpreted as the total probability (100 percent) that the physical system is found in any of the mutually excluding states, it must be true that the operator $H$ has to be Hermitian and, equivalently, $U$ has to be unitary or antiunitary. (Only the unitary option is possible if $U$ is obtained from $H$ by a continuous evolution.)

There doesn't exist a single experiment in which the linearity of quantum mechanics would be violated. In fact, by pure thought, one may see that such a deviation from linearity would require one to readjust the equations for $|\psi\rangle$ or their interpretations so that the total probability is returned back to 100 percent. Such readjustments would inevitably violate locality. Moreover, they would make the "collapse" of a wave function (and its precise moment) observable, at least in principle. These are de facto physical inconsistencies because they contradict the special theory of relativity.

At any rate, as of 2011, the postulates of quantum mechanics including linearity of the Hilbert space (of allowed wave functions) seem to be 100 percent valid and there doesn't even exist a logically plausible proposed way to "deform" them that wouldn't lead to other problems that are viewed as physical inconsistencies.

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If we assume time evolution preserves the Hilbert space norm, then Wigner had shown it can only be a linear unitary transformation, or an antilinear antiunitary transformation. If time evolution is continuous in time, it can't possibly be antiunitary as there's no continuous deformation of the identity operator to an antiunitary transformation.

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I don't know anything about Hilbert spaces. I guess that means that I have more reading to do –  Casebash Mar 26 '11 at 6:48

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