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Ground state

From Wikipedia, the free encyclopedia
Lowest energy level of a quantum system
Energy levels for anelectron in anatom:ground state andexcited states. After absorbingenergy, an electron mayjump from the ground state to a higher-energy excited state.

Theground state of aquantum-mechanical system is itsstationary state of lowestenergy; the energy of the ground state is known as thezero-point energy of the system. Anexcited state is any state with energy greater than the ground state. Inquantum field theory, the ground state is usually called 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 operator that acts non-trivially on a ground state andcommutes with theHamiltonian of the system.

According to thethird law of thermodynamics, a system atabsolute zerotemperature exists in its ground state; thus, itsentropy is 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 zero temperature for systems that exhibitnegative temperature.

Absence of nodes in one dimension

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In onedimension, the ground state of theSchrödinger equation can beproven to have nonodes.[1]

Derivation

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

⟨ψ|H|ψ⟩=∫dx(−ℏ22mψ∗d2ψdx2+V(x)|ψ(x)|2),{\displaystyle \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:

∫abψ∗d2ψdx2dx=[ψ∗dψdx]ab−∫abdψ∗dxdψdxdx=[ψ∗dψdx]ab−∫ab|dψdx|2dx{\displaystyle \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[ψ∗dψdx]−∞∞=limb→∞ψ∗(b)dψdx(b)−lima→−∞ψ∗(a)dψdx(a){\displaystyle \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:−ℏ22m∫−∞∞ψ∗d2ψdx2dx=ℏ22m∫−∞∞|dψdx|2dx{\displaystyle -{\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 smallinterval aroundx=0{\displaystyle x=0}; i.e.,x∈[−ε,ε]{\displaystyle x\in [-\varepsilon ,\varepsilon ]}. Take a new (deformed)wave functionψ'(x) to be defined asψ′(x)=ψ(x){\displaystyle \psi '(x)=\psi (x)}, forx<−ε{\displaystyle x<-\varepsilon }; andψ′(x)=−ψ(x){\displaystyle \psi '(x)=-\psi (x)}, forx>ε{\displaystyle x>\varepsilon }; andconstant forx∈[−ε,ε]{\displaystyle x\in [-\varepsilon ,\varepsilon ]}. Ifε{\displaystyle \varepsilon } is small enough, this is always possible to do, so thatψ'(x) is continuous.

Assumingψ(x)≈−cx{\displaystyle \psi (x)\approx -cx} aroundx=0{\displaystyle x=0}, one may writeψ′(x)=N{|ψ(x)|,|x|>ε,cε,|x|≤ε,{\displaystyle \psi '(x)=N{\begin{cases}|\psi (x)|,&|x|>\varepsilon ,\\c\varepsilon ,&|x|\leq \varepsilon ,\end{cases}}}whereN=11+43|c|2ε3{\displaystyle N={\frac {1}{\sqrt {1+{\frac {4}{3}}|c|^{2}\varepsilon ^{3}}}}} is the norm.

Note that the kinetic-energy densities holdℏ22m|dψ′dx|2<ℏ22m|dψdx|2{\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 energy is lowered byO(ε){\displaystyle O(\varepsilon )} by the deformation toψ'.

Now, consider thepotential energy. For definiteness, let us chooseV(x)≥0{\displaystyle V(x)\geq 0}. Then it is clear that, outside the intervalx∈[−ε,ε]{\displaystyle x\in [-\varepsilon ,\varepsilon ]}, the potential energy density is smaller for theψ' because|ψ′|<|ψ|{\displaystyle |\psi '|<|\psi |} there.

On the other hand, in the intervalx∈[−ε,ε]{\displaystyle x\in [-\varepsilon ,\varepsilon ]} we haveVavgε′=∫−εεdxV(x)|ψ′|2=ε2|c|21+43|c|2ε3∫−εεdxV(x)≃2ε3|c|2V(0)+⋯,{\displaystyle {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 orderε3{\displaystyle \varepsilon ^{3}}.

However, the contribution to the potential energy from this region for the stateψ with a node isVavgε=∫−εεdxV(x)|ψ|2=|c|2∫−εεdxx2V(x)≃23ε3|c|2V(0)+⋯,{\displaystyle 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 orderO(ε3){\displaystyle 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 orderε2{\displaystyle \varepsilon ^{2}}, if we deform the stateψ{\displaystyle \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 byO(ε){\displaystyle 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

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As the ground state has no nodes it isspatially non-degenerate, i.e. there are no twostationary quantum states with theenergy eigenvalue of the ground state (let's name itEg{\displaystyle E_{g}}) and the samespin state and therefore would only differ in their position-spacewave functions.[1]

The reasoning goes bycontradiction: For if the ground state would be degenerate then there would be two orthonormal[2] stationary states|ψ1⟩{\displaystyle \left|\psi _{1}\right\rangle } and|ψ2⟩{\displaystyle \left|\psi _{2}\right\rangle } — later on represented by their complex-valued position-space wave functionsψ1(x,t)=ψ1(x,0)⋅e−iEgt/ℏ{\displaystyle \psi _{1}(x,t)=\psi _{1}(x,0)\cdot e^{-iE_{g}t/\hbar }} andψ2(x,t)=ψ2(x,0)⋅e−iEgt/ℏ{\displaystyle \psi _{2}(x,t)=\psi _{2}(x,0)\cdot e^{-iE_{g}t/\hbar }} — and anysuperposition|ψ3⟩:=c1|ψ1⟩+c2|ψ2⟩{\displaystyle \left|\psi _{3}\right\rangle :=c_{1}\left|\psi _{1}\right\rangle +c_{2}\left|\psi _{2}\right\rangle } with the complex numbersc1,c2{\displaystyle c_{1},c_{2}} fulfilling the condition|c1|2+|c2|2=1{\displaystyle |c_{1}|^{2}+|c_{2}|^{2}=1} would also be a be such a state, i.e. would have the same energy-eigenvalueEg{\displaystyle E_{g}} and the same spin-state.

Now letx0{\displaystyle x_{0}} be some random point (where both wave functions are defined) and set:c1=ψ2(x0,0)a{\displaystyle c_{1}={\frac {\psi _{2}(x_{0},0)}{a}}}andc2=−ψ1(x0,0)a{\displaystyle c_{2}={\frac {-\psi _{1}(x_{0},0)}{a}}}witha=|ψ1(x0,0)|2+|ψ2(x0,0)|2>0{\displaystyle 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|ψ3⟩{\displaystyle \left|\psi _{3}\right\rangle } isψ3(x,t)=c1ψ1(x,t)+c2ψ2(x,t)=1a(ψ2(x0,0)⋅ψ1(x,0)−ψ1(x0,0)⋅ψ2(x,0))⋅e−iEgt/ℏ.{\displaystyle \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 }.}

Henceψ3(x0,t)=1a(ψ2(x0,0)⋅ψ1(x0,0)−ψ1(x0,0)⋅ψ2(x0,0))⋅e−iEgt/ℏ=0{\displaystyle \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 allt{\displaystyle t}.

But⟨ψ3|ψ3⟩=|c1|2+|c2|2=1{\displaystyle \left\langle \psi _{3}|\psi _{3}\right\rangle =|c_{1}|^{2}+|c_{2}|^{2}=1} i.e.,x0{\displaystyle x_{0}} isa 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 differentspin states like|↑⟩{\displaystyle \left|\uparrow \right\rangle } and|↓⟩{\displaystyle \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

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Initial wave functions for the first four states of a one-dimensional particle in a box
  • Thewave function of the ground state of aparticle in a one-dimensional box is a half-periodsine wave, which goes to zero at the two edges of the well. The energy of the particle is given byh2n28mL2{\textstyle {\frac {h^{2}n^{2}}{8mL^{2}}}}, whereh is thePlanck constant,m is the mass of the particle,n is the energy state (n = 1 corresponds to the ground-state energy), andL is 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 reducesexponentially at larger distances. Theelectron is 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 bebound to the atom.
  • The exact definition of onesecond oftime since 1997 has been the duration of9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of thecaesium-133 atom at rest at a temperature of 0 K.[3]

Notes

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  1. ^abSee, for example,Cohen, M. (1956)."Appendix A: Proof of non-degeneracy of the ground state"(PDF).The energy spectrum of the excitations in liquid helium (Ph.D.). California Institute of Technology. Published asFeynman, R. P.; Cohen, Michael (1956)."Energy Spectrum of the Excitations in Liquid Helium"(PDF).Physical Review.102 (5): 1189.Bibcode:1956PhRv..102.1189F.doi:10.1103/PhysRev.102.1189.
  2. ^i.e.⟨ψ1|ψ2⟩=δij{\displaystyle \left\langle \psi _{1}|\psi _{2}\right\rangle =\delta _{ij}}
  3. ^"Unit of time (second)".SI Brochure.International Bureau of Weights and Measures. Retrieved2013-12-22.

Bibliography

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