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Uncertainty principle
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===Matrix mechanics interpretation=== {{Main article|Matrix mechanics}} In matrix mechanics, observables such as position and momentum are represented by self-adjoint operators.<ref name="L&L"/> When considering pairs of observables, an important quantity is the ''[[commutator]]''. For a pair of operators {{mvar|Γ}} and <math>\hat{B}</math>, one defines their commutator as <math display="block">[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}.</math> In the case of position and momentum, the commutator is the [[canonical commutation relation]] <math display="block">[\hat{x},\hat{p}]=i \hbar.</math> The physical meaning of the non-commutativity can be understood by considering the effect of the commutator on position and momentum [[eigenstate]]s. Let <math>|\psi\rangle</math> be a right eigenstate of position with a constant eigenvalue {{math|''x''<sub>0</sub>}}. By definition, this means that <math>\hat{x}|\psi\rangle = x_0 |\psi\rangle.</math> Applying the commutator to <math>|\psi\rangle</math> yields <math display="block">[\hat{x},\hat{p}] | \psi \rangle = (\hat{x}\hat{p}-\hat{p}\hat{x}) | \psi \rangle = (\hat{x} - x_0 \hat{I}) \hat{p} \, | \psi \rangle = i \hbar | \psi \rangle,</math> where {{mvar|Γ}} is the [[identity matrix|identity operator]]. Suppose, for the sake of [[proof by contradiction]], that <math>|\psi\rangle</math> is also a right eigenstate of momentum, with constant eigenvalue {{mvar|''p''<sub>0</sub>}}. If this were true, then one could write <math display="block">(\hat{x} - x_0 \hat{I}) \hat{p} \, | \psi \rangle = (\hat{x} - x_0 \hat{I}) p_0 \, | \psi \rangle = (x_0 \hat{I} - x_0 \hat{I}) p_0 \, | \psi \rangle=0.</math> On the other hand, the above canonical commutation relation requires that <math display="block">[\hat{x},\hat{p}] | \psi \rangle=i \hbar | \psi \rangle \ne 0.</math> This implies that no quantum state can simultaneously be both a position and a momentum eigenstate. When a state is measured, it is projected onto an eigenstate in the basis of the relevant observable. For example, if a particle's position is measured, then the state amounts to a position eigenstate. This means that the state is ''not'' a momentum eigenstate, however, but rather it can be represented as a sum of multiple momentum basis eigenstates. In other words, the momentum must be less precise. This precision may be quantified by the standard deviations, <math display="block">\sigma_x=\sqrt{\langle \hat{x}^2 \rangle-\langle \hat{x}\rangle^2}</math> <math display="block">\sigma_p=\sqrt{\langle \hat{p}^2 \rangle-\langle \hat{p}\rangle^2}.</math> As in the wave mechanics interpretation above, one sees a tradeoff between the respective precisions of the two, quantified by the uncertainty principle.
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