Editing Ground state

{{Short description|Lowest energy level of a quantum system}}

[[File:Energy levels.svg|thumb|right|[[Energy level]]s for an [[electron]] in an [[atom]]: '''ground state''' and [[excited state]]s. After absorbing [[energy]], an electron may [[Atomic electron transition|jump]] from the ground state to a higher-energy excited state.]]

The '''ground state''' of a [[quantum mechanics|quantum-mechanical]] system is its [[stationary state]] of lowest [[energy]]; the energy of the ground state is known as the [[zero-point energy]] of the system. An [[excited state]] is any state with energy greater than the ground state. In [[quantum field theory]], the ground state is usually called the [[Quantum vacuum state|vacuum]].

If more than one ground state exists, they are said to be [[degenerate energy level|degenerate]]. Many systems have degenerate ground states. Degeneracy occurs whenever there exists a [[unitary operator]] that acts non-trivially on a ground state and [[commutator|commutes]] with the [[Hamiltonian (quantum mechanics)|Hamiltonian]] of the system.

According to the [[third law of thermodynamics]], a system at [[absolute zero]] [[temperature]] exists in its ground state; thus, its [[entropy]] is determined by the degeneracy of the ground state. Many systems, such as a perfect [[crystal lattice]], have a unique ground state and therefore have zero entropy at absolute zero. It is also possible for the highest excited state to have [[absolute zero]] temperature for systems that exhibit [[negative temperature]].

== Absence of nodes in one dimension ==
In one [[dimension]], the ground state of the [[Schrödinger equation]] can be [[Mathematical proof|proven]] to have no [[Node (physics)|nodes]].<ref name="Cohen">
See, for example, {{cite thesis |last= Cohen|first= M. |date=1956 |title=The energy spectrum of the excitations in liquid helium |type= Ph.D. |publisher=California Institute of Technology |chapter-url=https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf |chapter=Appendix A: Proof of non-degeneracy of the ground state}}  Published as {{cite journal |doi=10.1103/PhysRev.102.1189 |title=Energy Spectrum of the Excitations in Liquid Helium |journal=Physical Review |volume=102 |issue=5 |pages=1189 |year=1956 |last1=Feynman |first1=R. P. |last2=Cohen |first2=Michael |bibcode=1956PhRv..102.1189F|url=https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf }}</ref>

=== Derivation ===
Consider the [[Partition function (statistical mechanics)|average energy]] of a state with a node at {{math|1=''x'' = 0}}; i.e., {{math|1=''ψ''(0) = 0}}.  The average energy in this state would be

<math display="block"> \langle\psi|H|\psi\rangle = \int dx\, \left(-\frac{\hbar^2}{2m} \psi^* \frac{d^2\psi}{dx^2} + V(x)|\psi(x)|^2\right),</math>

where {{math|''V''(''x'')}} is the potential.

With [[integration by parts]]:

<math display="block">\int_a^b \psi^* \frac{d^2\psi}{dx^2} dx
= \left[ \psi^*\frac{d\psi}{dx}\right]_a^b - \int_a^b \frac{d\psi^*}{dx}\frac{d\psi}{dx} dx
=  \left[ \psi^*\frac{d\psi}{dx}\right]_a^b - \int_a^b \left|\frac{d\psi}{dx}\right|^2 dx </math>

Hence in case that <math>\left[ \psi^*\frac{d\psi}{dx}\right]_{-\infty}^{\infty} = \lim_{b\to\infty}\psi^*(b)\frac{d\psi}{dx}(b)-\lim_{a\to-\infty}\psi^*(a)\frac{d\psi}{dx}(a)</math> is equal to ''zero'', one gets:
<math display="block">-\frac{\hbar^2}{2m}\int_{-\infty}^{\infty} \psi^* \frac{d^2\psi}{dx^2} dx
= \frac{\hbar^2}{2m}\int_{-\infty}^{\infty} \left|\frac{d\psi}{dx}\right|^2 dx</math>

Now, consider a small [[Interval (mathematics)|interval]] around <math>x = 0</math>; i.e., <math>x \in [-\varepsilon, \varepsilon]</math>. Take a new ([[Deformation (mathematics)|deformed]]) [[wave function]] {{math|''ψ{{'}}''(''x'')}} to be defined as <math>\psi'(x) = \psi(x)</math>, for <math>x < -\varepsilon</math>; and <math>\psi'(x) = -\psi(x)</math>, for <math>x > \varepsilon</math>; and [[Constant (mathematics)|constant]] for <math>x \in [-\varepsilon, \varepsilon]</math>. If <math>\varepsilon</math> is small enough, this is always possible to do, so that  {{math|''ψ{{'}}''(''x'')}}  is continuous.

Assuming <math>\psi(x) \approx -cx</math> around <math>x = 0</math>, one may write
<math display="block">
  \psi'(x) = N \begin{cases}
    |\psi(x)|,    & |x| >   \varepsilon, \\
    c\varepsilon, & |x| \le \varepsilon,
  \end{cases}
</math>
where <math>N = \frac{1}{\sqrt{1 + \frac{4}{3} |c|^2\varepsilon^3}}</math> is the norm.

Note that the kinetic-energy densities hold <math display="inline">\frac{\hbar^2}{2m}\left|\frac{d\psi'}{dx}\right|^2 < \frac{\hbar^2}{2m}\left|\frac{d\psi}{dx}\right|^2</math> everywhere because of the normalization. More significantly, the average [[kinetic energy]] is lowered by <math>O(\varepsilon)</math> by the deformation to {{math|''ψ{{'}}''}}.

Now, consider the [[potential energy]]. For definiteness, let us choose <math>V(x) \ge 0</math>. Then it is clear that, outside the interval <math>x \in [-\varepsilon, \varepsilon]</math>, the potential energy density is smaller for the {{math|''ψ{{'}}''}} because <math>|\psi'| < |\psi|</math> there.

On the other hand, in the interval <math>x \in [-\varepsilon, \varepsilon]</math> we have
<math display="block">
  {V^\varepsilon_\text{avg}}' = \int_{-\varepsilon}^\varepsilon dx\, V(x)|\psi'|^2 = \frac{\varepsilon^2|c|^2}{1 + \frac{4}{3}|c|^2\varepsilon^3} \int_{-\varepsilon}^\varepsilon dx\, V(x) \simeq 2\varepsilon^3|c|^2 V(0) + \cdots,
</math>
which holds to order <math>\varepsilon^3</math>.

However, the contribution to the potential energy from this region for the state {{math|''ψ''}} with a node is
<math display="block">
  V^\varepsilon_\text{avg} = \int_{-\varepsilon}^\varepsilon dx\, V(x)|\psi|^2 = |c|^2\int_{-\varepsilon}^\varepsilon dx\, x^2V(x) \simeq \frac{2}{3}\varepsilon^3|c|^2 V(0) + \cdots,
</math>
lower, but still of the same lower order <math>O(\varepsilon^3)</math> as for the deformed state  {{math|''ψ{{'}}''}}, and subdominant to the lowering of the average kinetic energy.
Therefore, the potential energy is unchanged up to order <math>\varepsilon^2</math>, if we deform the state <math>\psi</math> with a node into a state {{math|''ψ{{'}}''}} without a node, and the change can be ignored.

We can therefore remove all nodes and reduce the energy by <math>O(\varepsilon)</math>, which implies that {{math|''ψ{{'}}''}} cannot be the ground state. Thus the ground-state wave function cannot have a node. This completes the proof. (The average energy may then be further lowered by eliminating undulations, to the variational absolute minimum.)

=== Implication ===
As the ground state has no nodes it is ''spatially'' non-degenerate, i.e. there are no two [[Stationary state|stationary quantum states]] with the [[Hamiltonian (quantum mechanics)|energy eigenvalue]] of the ground state (let's name it <math>E_g</math>) and the same [[Spin (physics)|spin state]] and therefore would only differ in their position-space [[wave function]]s.<ref name="Cohen"/>

The reasoning goes by [[Proof by contradiction|contradiction]]: For if the ground state would be degenerate then there would be two orthonormal<ref>i.e. <math>\left\lang \psi_1|\psi_2\right\rang = \delta_{ij}</math></ref> stationary states <math>\left|\psi_1\right\rang</math> and <math>\left|\psi_2\right\rang</math> — later on represented by their complex-valued position-space wave functions <math>\psi_1(x,t)=\psi_1(x,0)\cdot e^{-iE_g t/\hbar}</math> and <math>\psi_2(x,t)=\psi_2(x,0)\cdot e^{-iE_g t/\hbar}</math> — and any [[Quantum superposition|superposition]] <math>\left|\psi_3\right\rang := c_1\left|\psi_1\right\rang + c_2\left|\psi_2\right\rang</math> with the complex numbers <math>c_1, c_2</math> fulfilling the condition <math>|c_1|^2+|c_2|^2=1</math> would also be a be such a state, i.e. would have the same energy-eigenvalue <math>E_g</math> and the same spin-state.

Now let <math>x_0</math> be some random point (where both wave functions are defined) and set:
<math display="block">c_1=\frac{\psi_2(x_0,0)}{a}</math>
and
<math display="block">c_2=\frac{-\psi_1(x_0,0)}{a}</math>
with
<math display="block">a=\sqrt{|\psi_1(x_0,0)|^2+|\psi_2(x_0,0)|^2} > 0</math>
(according to the premise ''no nodes'').

Therefore, the position-space wave function of <math>\left|\psi_3\right\rang</math> is
<math display="block">\psi_3(x,t)=c_1\psi_1(x,t)+c_2\psi_2(x,t) = \frac{1}{a}\left(\psi_2(x_0,0)\cdot\psi_1(x,0) - \psi_1(x_0,0)\cdot\psi_2(x,0) \right)\cdot e^{-iE_g t/\hbar}.</math>

Hence
<math display="block">\psi_3(x_0,t)=\frac{1}{a}\left(\psi_2(x_0,0)\cdot\psi_1(x_0,0) - \psi_1(x_0,0)\cdot\psi_2(x_0,0) \right)\cdot e^{-iE_g t/\hbar} = 0 </math>
for all <math>t</math>.

But <math>\left\lang \psi_3|\psi_3\right\rang = |c_1|^2+|c_2|^2=1</math> i.e., <math>x_0</math> is ''a node'' of the ground state wave function and that is in contradiction to the premise that this wave function cannot have a node.

Note that the ground state could be degenerate because of different ''spin states'' like <math>\left|\uparrow\right\rang</math> and <math>\left|\downarrow\right\rang</math> while having the same position-space wave function: Any superposition of these states would create a mixed spin state but leave the spatial part (as a common factor of both) unaltered.

== Examples ==
[[File:particle in a box wavefunctions 2.svg|thumb|right|upright|Initial wave functions for the first four states of a one-dimensional particle in a box]]
* The [[wave function]] of the ground state of a [[particle in a box|particle in a one-dimensional box]] is a half-period [[sine wave]], which goes to zero at the two edges of the well. The energy of the particle is given by <math display="inline">\frac{h^2 n^2}{8 m L^2}</math>, where ''h'' is the [[Planck constant]], ''m'' is the mass of the particle, ''n'' is the energy state (''n'' = 1 corresponds to the ground-state energy), and ''L'' is the width of the well.
* The wave function of the ground state of a hydrogen atom is a spherically symmetric distribution centred on the [[atomic nucleus|nucleus]], which is largest at the center and reduces [[exponential distribution|exponentially]] at larger distances. The [[electron]] is most likely to be found at a distance from the nucleus equal to the [[Bohr radius]]. This function is known as the 1s [[atomic orbital]]. For hydrogen (H), an electron in the ground state has energy {{val|-13.6|u=eV}}, relative to the [[Ionization energy|ionization threshold]]. In other words, 13.6&nbsp;eV is the energy input required for the electron to no longer be [[Bound state|bound]] to the atom.
* The exact definition of one [[second]] of [[time]] since 1997 has been the duration of {{val|9,192,631,770}} periods of the radiation corresponding to the transition between the two [[hyperfine structure | hyperfine]] levels of the ground state of the [[caesium]]-133 atom at rest at a temperature of 0&nbsp;K.<ref>{{cite web
 |title=Unit of time (second)
 |work=SI Brochure
 |url=http://www.bipm.org/en/si/si_brochure/chapter2/2-1/second.html
 |publisher=[[International Bureau of Weights and Measures]]
 |access-date=2013-12-22
}}</ref>

== Notes ==
{{Reflist}}

== Bibliography ==
* {{cite book|title=The Feynman Lectures on Physics|volume=3| first1=Richard |last1=Feynman|author1-link=Richard Feynman|first2=Robert |last2=Leighton|first3=Matthew|last3= Sands|year=1965|chapter=see section 2-5 for energy levels, 19 for the hydrogen atom|chapter-url=https://feynmanlectures.caltech.edu/III_toc.html}}
<!-- Covered in any QM textbook – from the Feynman lectures, see section 2-5 for energy levels, 19 for the hydrogen atom -->

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[[Category:Quantum states]]