Ground state

Energy levelsfor anelectronin anatom:ground stateandexcited states.After absorbingenergy,an electron mayjumpfrom the ground state to a higher-energy excited state.

Theground stateof aquantum-mechanicalsystem is itsstationary stateof lowestenergy;the energy of the ground state is known as thezero-point energyof the system. Anexcited stateis any state with energy greater than the ground state. Inquantum field theory,the ground state is usually called thevacuum stateor thevacuum.

If more than one ground state exists, they are said to bedegenerate.Many systems have degenerate ground states. Degeneracy occurs whenever there exists aunitary operatorthat acts non-trivially on a ground state andcommuteswith theHamiltonianof the system.

According to thethird law of thermodynamics,a system atabsolute zerotemperatureexists in its ground state; thus, itsentropyis determined by the degeneracy of the ground state. Many systems, such as a perfectcrystal lattice,have a unique ground state and therefore have zero entropy at absolute zero. It is also possible for the highest excited state to haveabsolute zerotemperature for systems that exhibitnegative temperature.

Absence of nodes in one dimension

In onedimension,the ground state of theSchrödinger equationcan beprovento have nonodes.

Derivation

Consider theaverage energyof a state with a node atx= 0;i.e.,ψ(0) = 0.The average energy in this state would be

$$\langle \psi |H|\psi \rangle =\int dx\,\left(-{\frac {\hbar ^{2}}{2m}}\psi ^{*}{\frac {d^{2}\psi }{dx^{2}}}+V(x)|\psi (x)|^{2}\right),$$

whereV(x)is the potential.

Withintegration by parts: $$\int _{a}^{b}\psi ^{*}{\frac {d^{2}\psi }{dx^{2}}}dx=\left[\psi ^{*}{\frac {d\psi }{dx}}\right]_{a}^{b}-\int _{a}^{b}{\frac {d\psi ^{*}}{dx}}{\frac {d\psi }{dx}}dx=\left[\psi ^{*}{\frac {d\psi }{dx}}\right]_{a}^{b}-\int _{a}^{b}\left|{\frac {d\psi }{dx}}\right|^{2}dx$$

Hence in case that$\left[\psi ^{*}{\frac {d\psi }{dx}}\right]_{-\infty }^{\infty }=\lim _{b\to \infty }\psi ^{*}(b){\frac {d\psi }{dx}}(b)-\lim _{a\to -\infty }\psi ^{*}(a){\frac {d\psi }{dx}}(a)$is equal tozero,one gets: $$-{\frac {\hbar ^{2}}{2m}}\int _{-\infty }^{\infty }\psi ^{*}{\frac {d^{2}\psi }{dx^{2}}}dx={\frac {\hbar ^{2}}{2m}}\int _{-\infty }^{\infty }\left|{\frac {d\psi }{dx}}\right|^{2}dx$$

Now, consider a smallintervalaround$x=0$;i.e.,$x\in [-\varepsilon,\varepsilon ]$.Take a new (deformed)wave functionψ'(x)to be defined as$\psi '(x)=\psi (x)$,for$x<-\varepsilon$;and$\psi '(x)=-\psi (x)$,for$x>\varepsilon$;andconstantfor$x\in [-\varepsilon,\varepsilon ]$.If$\varepsilon$is small enough, this is always possible to do, so thatψ'(x)is continuous.

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

Note that the kinetic-energy densities hold${\textstyle {\frac {\hbar ^{2}}{2m}}\left|{\frac {d\psi '}{dx}}\right|^{2}<{\frac {\hbar ^{2}}{2m}}\left|{\frac {d\psi }{dx}}\right|^{2}}$everywhere because of the normalization. More significantly, the averagekinetic energyis lowered by$O(\varepsilon )$by the deformation toψ'.

Now, consider thepotential energy.For definiteness, let us choose$V(x)\geq 0$.Then it is clear that, outside the interval$x\in [-\varepsilon,\varepsilon ]$,the potential energy density is smaller for theψ'because$|\psi '|<|\psi |$there.

On the other hand, in the interval$x\in [-\varepsilon,\varepsilon ]$we have $${V_{\text{avg}}^{\varepsilon }}'=\int _{-\varepsilon }^{\varepsilon }dx\,V(x)|\psi '|^{2}={\frac {\varepsilon ^{2}|c|^{2}}{1+{\frac {4}{3}}|c|^{2}\varepsilon ^{3}}}\int _{-\varepsilon }^{\varepsilon }dx\,V(x)\simeq 2\varepsilon ^{3}|c|^{2}V(0)+\cdots,$$ which holds to order$\varepsilon ^{3}$.

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

We can therefore remove all nodes and reduce the energy by$O(\varepsilon )$,which implies thatψ'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 isspatiallynon-degenerate, i.e. there are no twostationary quantum stateswith theenergy eigenvalueof the ground state (let's name it$E_{g}$) and the samespin stateand therefore would only differ in their position-spacewave functions.

The reasoning goes bycontradiction:For if the ground state would be degenerate then there would be two orthonormalstationary states$\left|\psi _{1}\right\rangle$and$\left|\psi _{2}\right\rangle$— later on represented by their complex-valued position-space wave functions$\psi _{1}(x,t)=\psi _{1}(x,0)\cdot e^{-iE_{g}t/\hbar }$and$\psi _{2}(x,t)=\psi _{2}(x,0)\cdot e^{-iE_{g}t/\hbar }$— and anysuperposition$\left|\psi _{3}\right\rangle:=c_{1}\left|\psi _{1}\right\rangle +c_{2}\left|\psi _{2}\right\rangle$with the complex numbers$c_{1},c_{2}$fulfilling the condition$|c_{1}|^{2}+|c_{2}|^{2}=1$would also be a be such a state, i.e. would have the same energy-eigenvalue$E_{g}$and the same spin-state.

Now let$x_{0}$be some random point (where both wave functions are defined) and set: $$c_{1}={\frac {\psi _{2}(x_{0},0)}{a}}$$ and $$c_{2}={\frac {-\psi _{1}(x_{0},0)}{a}}$$ with $$a={\sqrt {|\psi _{1}(x_{0},0)|^{2}+|\psi _{2}(x_{0},0)|^{2}}}>0$$ (according to the premiseno nodes).

Therefore, the position-space wave function of$\left|\psi _{3}\right\rangle$is $$\psi _{3}(x,t)=c_{1}\psi _{1}(x,t)+c_{2}\psi _{2}(x,t)={\frac {1}{a}}\left(\psi _{2}(x_{0},0)\cdot \psi _{1}(x,0)-\psi _{1}(x_{0},0)\cdot \psi _{2}(x,0)\right)\cdot e^{-iE_{g}t/\hbar }.$$

Hence $$\psi _{3}(x_{0},t)={\frac {1}{a}}\left(\psi _{2}(x_{0},0)\cdot \psi _{1}(x_{0},0)-\psi _{1}(x_{0},0)\cdot \psi _{2}(x_{0},0)\right)\cdot e^{-iE_{g}t/\hbar }=0$$ for all$t$.

But$\left\langle \psi _{3}|\psi _{3}\right\rangle =|c_{1}|^{2}+|c_{2}|^{2}=1$i.e.,$x_{0}$isa nodeof 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 differentspin stateslike$\left|\uparrow \right\rangle$and$\left|\downarrow \right\rangle$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

Initial wave functions for the first four states of a one-dimensional particle in a box

• Thewave functionof the ground state of aparticle in a one-dimensional boxis a half-periodsine wave,which goes to zero at the two edges of the well. The energy of the particle is given by${\textstyle {\frac {h^{2}n^{2}}{8mL^{2}}}}$,wherehis thePlanck constant,mis the mass of the particle,nis the energy state (n= 1 corresponds to the ground-state energy), andLis the width of the well.
• The wave function of the ground state of a hydrogen atom is a spherically symmetric distribution centred on thenucleus,which is largest at the center and reducesexponentiallyat larger distances. Theelectronis most likely to be found at a distance from the nucleus equal to theBohr radius.This function is known as the 1satomic orbital.For hydrogen (H), an electron in the ground state has energy−13.6 eV,relative to theionization threshold.In other words, 13.6 eV is the energy input required for the electron to no longer beboundto the atom.
• The exact definition of onesecondoftimesince 1997 has been the duration of9192631770periods of the radiation corresponding to the transition between the twohyperfinelevels of the ground state of thecaesium-133 atom at rest at a temperature of 0 K.

Bibliography

• Feynman, Richard;Leighton, Robert; Sands, Matthew (1965)."see section 2-5 for energy levels, 19 for the hydrogen atom".The Feynman Lectures on Physics.Vol. 3.