Editing Zero-point energy (section)

== Uncertainty principle ==
{{Main|Uncertainty principle}}

Zero-point energy is fundamentally related to the Heisenberg uncertainty principle.<ref name="Heisenberg_1927">{{Cite journal |last=Heisenberg |first=Werner |year=1927 |title=Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik |trans-title=On the evident content of quantum-theoretical kinematics and mechanics |url=http://scarc.library.oregonstate.edu/coll/pauling/bond/papers/corr155.1.html |journal=Zeitschrift für Physik |language=de |volume=43 |issue=3 |pages=172–198 |bibcode=1927ZPhy...43..172H |doi=10.1007/BF01397280 |s2cid=122763326|url-access=subscription }}</ref> Roughly speaking, the uncertainty principle states that complementary variables (such as a particle's position and [[momentum]], or a field's value and derivative at a point in space) cannot simultaneously be specified precisely by any given quantum state. In particular, there cannot exist a state in which the system simply sits motionless at the bottom of its potential well, for then its position and momentum would both be completely determined to arbitrarily great precision. Therefore, the lowest-energy state (the ground state) of the system must have a distribution in position and momentum that satisfies the uncertainty principle, which implies its energy must be greater than the minimum of the potential well.

Near the bottom of a [[potential well]], the [[Hamiltonian (quantum mechanics)|Hamiltonian]] of a general system (the quantum-mechanical [[operator (physics)|operator]] giving its energy) can be approximated as a [[quantum harmonic oscillator]],
<math display="block">\hat{H} = V_0 + \tfrac{1}{2} k \left(\hat{x} - x_0\right)^2 + \frac{1}{2m} \hat{p}^2 \,,</math>
where {{math|''V''<sub>0</sub>}} is the minimum of the classical potential well.

The uncertainty principle tells us that
<math display="block">\sqrt{\left\langle \left(\hat{x} - x_0\right)^2 \right\rangle} \sqrt{\left\langle \hat{p}^2 \right\rangle} \geq \frac{\hbar}{2} \,,</math>
making the [[expectation value (quantum mechanics)|expectation value]]s of the kinetic and [[potential energy|potential]] terms above satisfy
<math display="block">\left\langle \tfrac{1}{2} k \left(\hat{x} - x_0\right)^2 \right\rangle \left\langle \frac{1}{2m} \hat{p}^2 \right\rangle \geq \left(\frac{\hbar}{4}\right)^2 \frac{k}{m} \,.</math>

The expectation value of the energy must therefore be at least

<math display="block">\left\langle \hat{H} \right\rangle \geq V_0 + \frac{\hbar}{2} \sqrt{\frac{k}{m}} = V_0 + \frac{\hbar \omega}{2}</math>

where {{math|''ω'' {{=}} {{sqrt|''k''/''m''}}}} is the [[angular frequency]] at which the system oscillates.

A more thorough treatment, showing that the energy of the ground state actually saturates this bound and is exactly {{math|''E''<sub>0</sub> {{=}} ''V''<sub>0</sub> + {{sfrac|''ħω''|2}}}}, requires solving for the ground state of the system.