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What can be the precise answer to the question that

Quantum states are complex and infinite dimensional. Why is this so?

Is it because they belong to the complex Hilbert space? Even if they belong to the Hilbert space, why they are infinite dimensional, is this because of the projection operator?

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    $\begingroup$ You realize that "Quantum states are complex and infinite dimensional" is not a question, just a statement. Is your question "Why are quantum states infinite dimensional?" $\endgroup$ – jhobbie Jul 31 '14 at 14:58
  • $\begingroup$ My question is both- Why they are complex as well as why they are infinite dimensional ? $\endgroup$ – Roshan Shrestha Jul 31 '14 at 14:59
  • $\begingroup$ The question about QM and complex numbers has been asked before: physics.stackexchange.com/q/8062/2451 and links therein. $\endgroup$ – Qmechanic Jul 31 '14 at 15:11
  • $\begingroup$ Unfortunately I am still not get the precise answer, though I do think that there are not any. Also, what about why they are infinite dimensional. $\endgroup$ – Roshan Shrestha Jul 31 '14 at 15:23
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    $\begingroup$ The statement is wrong. The states are the one dimensional subspaces of the corresponding Hilbert space. $\endgroup$ – MBN Jul 31 '14 at 15:28
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The canonical quantization procedure prescribes that

$$ [x,p] = \mathrm{i\hbar}$$

Now, take the trace on both sides. The trace of a commutator vanishes, and, if the Hilbert space $\mathcal{H}$ were finite-dimensional, we get

$$ 0 = \mathrm{i}\hbar\mathrm{dim}(\mathcal{H}) $$

which is obviously false, so the assumption of finite dimensionality is wrong. Therefore, the trace must not be defined on the identity of $\mathcal{H}$, and we conclude that Hilbert spaces obtained by canoncial quantization are infinite-dimensional.

Note that not all QM Hilbert spaces are infinite. The Hilbert space of a non-moving spin-$\frac{1}{2}$ particle, for example, is only two dimensional.

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    $\begingroup$ He also asked why they are complex. Again, the commutator of two Hermitian observables like $x,p$ is anti-Hermitian, and if that's of the form $i\hbar$, it has to have non-real complex entries on the diagonal. It follows that either $x$ or $p$ or both have to have some complex entries, too, otherwise the commutator would be real matrix. Complex numbers are also necessary because $i$ expplicitly appears in Schrodinger's, Heisenerg's, or Dirac's equations of motion, or in Feynman path integrals - in all dynamical evolution equations in QM in any picture. $\endgroup$ – Luboš Motl Jul 31 '14 at 16:08
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Quantum states are not always complex and infinite dimensional. For example state which represents only spin of spin-1/2 particle is 2-dimensional and in matrix representation can be written for example as (1,0) for "spin up" and (0,1) for "spin down". They are complex because in general instead of ones and zeros there can be any complex numbers (up to normalization). Infinite dimensional state vectors arise when we want to describe, for example, position of a particle because space has infinitely many points. Each point has to have basis vector associated with it and general quantum state of that particle is sum of this (infinitely many) basis vectors with some (complex) coefficients.

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