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Qmechanic
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Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators UncertaintyOperators Uncertainty

Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

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Qmechanic
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Heisenberg's uncertainty principle states that: if the x-component of the momentum of a particle is measured with an uncertainty

$$\Delta \vec p_x$$

then its x-position cannot, at same time, be measured more accurately than $$\Delta\vec x=\frac {\hbar}{2\Delta\vec p_x},$$

$$\Delta\vec x\Delta\vec p_x \ge \frac {\hbar}{2}.$$$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

Heisenberg's uncertainty principle states that: if the x-component of the momentum of a particle is measured with an uncertainty

$$\Delta \vec p_x$$

then its x-position cannot, at same time, be measured more accurately than $$\Delta\vec x=\frac {\hbar}{2\Delta\vec p_x},$$

$$\Delta\vec x\Delta\vec p_x \ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

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went back to original user8784 question. The previous change was unwarranted.
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anna v
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changed "scientific" to "experimental" after a comment from the OP
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Sklivvz
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[Operators Uncertainty](http://physics.stackexchange.com/questions/24178/operators-uncertainty)
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Qmechanic
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layout; retagged; added wiki link;
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Qmechanic
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Post Migrated Here from theoreticalphysics.stackexchange.com (revisions)
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