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Ground state
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<h2 id="absence-of-nodes-in-one-dimension">Absence of nodes in one dimension</h2> <p>In one <a href="/facts/Dimension/0ktmUyac">dimension</a>, the ground state of the <a href="/facts/Schr%25C3%25B6dinger_equation/3csgSIWF">Schrödinger equation</a> can be <a href="/facts/Mathematical_proof/7ECDrU80">proven</a> to have no <a href="/facts/Node_(physics)/7tE2rDnb">nodes</a>.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> <h3>Derivation</h3> <p>Consider the <a href="/facts/Partition_function_(statistical_mechanics)/q65sJrrn">average energy</a> of a state with a node at <i>x</i> = 0; i.e., <i>ψ</i>(0) = 0. The average energy in this state would be </p><p> ⟨ ψ | H | ψ ⟩ = ∫ d x ( − ℏ 2 2 m ψ ∗ d 2 ψ d x 2 + 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),} </p><p>where <i>V</i>(<i>x</i>) is the potential. </p><p>With <a href="/facts/Integration_by_parts/aV6JgBNd">integration by parts</a>: </p><p> ∫ a b ψ ∗ d 2 ψ d x 2 d x = [ ψ ∗ d ψ d x ] a b − ∫ a b d ψ ∗ d x d ψ d x d x = [ ψ ∗ d ψ d x ] a b − ∫ a b | d ψ d x | 2 d x {\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} </p><p>Hence in case that [ ψ ∗ d ψ d x ] − ∞ ∞ = lim b → ∞ ψ ∗ ( b ) d ψ d x ( b ) − lim a → − ∞ ψ ∗ ( a ) d ψ d x ( 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 to <i>zero</i>, one gets: − ℏ 2 2 m ∫ − ∞ ∞ ψ ∗ d 2 ψ d x 2 d x = ℏ 2 2 m ∫ − ∞ ∞ | d ψ d x | 2 d x {\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} </p><p>Now, consider a small <a href="/facts/Interval_(mathematics)/e9se3vTe">interval</a> around x = 0 {\displaystyle x=0} ; i.e., x ∈ [ − ε , ε ] {\displaystyle x\in [-\varepsilon ,\varepsilon ]} . Take a new (<a href="/facts/Deformation_(mathematics)/tVLS3kkG">deformed</a>) <a href="/facts/Wave_function/KgM5ykjY">wave function</a> <i>ψ'</i>(<i>x</i>) to be defined as ψ ′ ( x ) = ψ ( x ) {\displaystyle \psi '(x)=\psi (x)} , for x < − ε {\displaystyle x<-\varepsilon } ; and ψ ′ ( x ) = − ψ ( x ) {\displaystyle \psi '(x)=-\psi (x)} , for x > ε {\displaystyle x>\varepsilon } ; and <a href="/facts/Constant_(mathematics)/SxlxTWFT">constant</a> for x ∈ [ − ε , ε ] {\displaystyle x\in [-\varepsilon ,\varepsilon ]} . If ε {\displaystyle \varepsilon } is small enough, this is always possible to do, so that <i>ψ'</i>(<i>x</i>) is continuous. </p><p>Assuming ψ ( x ) ≈ − c x {\displaystyle \psi (x)\approx -cx} around x = 0 {\displaystyle x=0} , one may write ψ ′ ( x ) = N { | ψ ( x ) | , | x | > ε , c ε , | x | ≤ ε , {\displaystyle \psi '(x)=N{\begin{cases}|\psi (x)|,&|x|>\varepsilon ,\\c\varepsilon ,&|x|\leq \varepsilon ,\end{cases}}} where N = 1 1 + 4 3 | c | 2 ε 3 {\displaystyle N={\frac {1}{\sqrt {1+{\frac {4}{3}}|c|^{2}\varepsilon ^{3}}}}} is the norm. </p><p>Note that the kinetic-energy densities hold ℏ 2 2 m | d ψ ′ d x | 2 < ℏ 2 2 m | d ψ d x | 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 average <a href="/facts/Kinetic_energy/aJRxUd4p">kinetic energy</a> is lowered by O ( ε ) {\displaystyle O(\varepsilon )} by the deformation to <i>ψ'</i>. </p><p>Now, consider the <a href="/facts/Potential_energy/I03KLbmn">potential energy</a>. For definiteness, let us choose V ( x ) ≥ 0 {\displaystyle V(x)\geq 0} . Then it is clear that, outside the interval x ∈ [ − ε , ε ] {\displaystyle x\in [-\varepsilon ,\varepsilon ]} , the potential energy density is smaller for the <i>ψ'</i> because | ψ ′ | < | ψ | {\displaystyle |\psi '|<|\psi |} there. </p><p>On the other hand, in the interval x ∈ [ − ε , ε ] {\displaystyle x\in [-\varepsilon ,\varepsilon ]} we have V avg ε ′ = ∫ − ε ε d x V ( x ) | ψ ′ | 2 = ε 2 | c | 2 1 + 4 3 | c | 2 ε 3 ∫ − ε ε d x V ( x ) ≃ 2 ε 3 | c | 2 V ( 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 order ε 3 {\displaystyle \varepsilon ^{3}} . </p><p>However, the contribution to the potential energy from this region for the state <i>ψ</i> with a node is V avg ε = ∫ − ε ε d x V ( x ) | ψ | 2 = | c | 2 ∫ − ε ε d x x 2 V ( x ) ≃ 2 3 ε 3 | c | 2 V ( 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 order O ( ε 3 ) {\displaystyle O(\varepsilon ^{3})} as for the deformed state <i>ψ'</i>, and subdominant to the lowering of the average kinetic energy. Therefore, the potential energy is unchanged up to order ε 2 {\displaystyle \varepsilon ^{2}} , if we deform the state ψ {\displaystyle \psi } with a node into a state <i>ψ'</i> without a node, and the change can be ignored. </p><p>We can therefore remove all nodes and reduce the energy by O ( ε ) {\displaystyle O(\varepsilon )} , which implies that <i>ψ'</i> 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.) </p> <h3>Implication</h3> <p>As the ground state has no nodes it is <i>spatially</i> non-degenerate, i.e. there are no two <a href="/facts/Stationary_state/HBQvbb6s">stationary quantum states</a> with the <a href="/facts/Hamiltonian_(quantum_mechanics)/5XwK2HmD">energy eigenvalue</a> of the ground state (let's name it E g {\displaystyle E_{g}} ) and the same <a href="/facts/Spin_(physics)/4Jvp7e8P">spin state</a> and therefore would only differ in their position-space <a href="/facts/Wave_function/KgM5ykjY">wave functions</a>.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p><p>The reasoning goes by <a href="/facts/Proof_by_contradiction/vHtFzFQ5">contradiction</a>: For if the ground state would be degenerate then there would be two orthonormal<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> stationary states | ψ 1 ⟩ {\displaystyle \left|\psi _{1}\right\rangle } and | ψ 2 ⟩ {\displaystyle \left|\psi _{2}\right\rangle } — later on represented by their complex-valued position-space wave functions ψ 1 ( x , t ) = ψ 1 ( x , 0 ) ⋅ e − i E g t / ℏ {\displaystyle \psi _{1}(x,t)=\psi _{1}(x,0)\cdot e^{-iE_{g}t/\hbar }} and ψ 2 ( x , t ) = ψ 2 ( x , 0 ) ⋅ e − i E g t / ℏ {\displaystyle \psi _{2}(x,t)=\psi _{2}(x,0)\cdot e^{-iE_{g}t/\hbar }} — and any <a href="/facts/Quantum_superposition/4z1UJg9o">superposition</a> | ψ 3 ⟩ := c 1 | ψ 1 ⟩ + c 2 | ψ 2 ⟩ {\displaystyle \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 {\displaystyle c_{1},c_{2}} fulfilling the condition | c 1 | 2 + | c 2 | 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-eigenvalue E g {\displaystyle E_{g}} and the same spin-state. </p><p>Now let x 0 {\displaystyle x_{0}} be some random point (where both wave functions are defined) and set: c 1 = ψ 2 ( x 0 , 0 ) a {\displaystyle c_{1}={\frac {\psi _{2}(x_{0},0)}{a}}} and c 2 = − ψ 1 ( x 0 , 0 ) a {\displaystyle c_{2}={\frac {-\psi _{1}(x_{0},0)}{a}}} with a = | ψ 1 ( x 0 , 0 ) | 2 + | ψ 2 ( x 0 , 0 ) | 2 > 0 {\displaystyle a={\sqrt {|\psi _{1}(x_{0},0)|^{2}+|\psi _{2}(x_{0},0)|^{2}}}>0} (according to the premise <i>no nodes</i>). </p><p>Therefore, the position-space wave function of | ψ 3 ⟩ {\displaystyle \left|\psi _{3}\right\rangle } is ψ 3 ( x , t ) = c 1 ψ 1 ( x , t ) + c 2 ψ 2 ( x , t ) = 1 a ( ψ 2 ( x 0 , 0 ) ⋅ ψ 1 ( x , 0 ) − ψ 1 ( x 0 , 0 ) ⋅ ψ 2 ( x , 0 ) ) ⋅ e − i E g t / 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 }.} </p><p>Hence ψ 3 ( x 0 , t ) = 1 a ( ψ 2 ( x 0 , 0 ) ⋅ ψ 1 ( x 0 , 0 ) − ψ 1 ( x 0 , 0 ) ⋅ ψ 2 ( x 0 , 0 ) ) ⋅ e − i E g t / 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 all t {\displaystyle t} . </p><p>But ⟨ ψ 3 | ψ 3 ⟩ = | c 1 | 2 + | c 2 | 2 = 1 {\displaystyle \left\langle \psi _{3}|\psi _{3}\right\rangle =|c_{1}|^{2}+|c_{2}|^{2}=1} i.e., x 0 {\displaystyle x_{0}} is <i>a node</i> of the ground state wave function and that is in contradiction to the premise that this wave function cannot have a node. </p><p>Note that the ground state could be degenerate because of different <i>spin states</i> 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. </p> <h2 id="examples">Examples</h2> <ul><li>The <a href="/facts/Wave_function/KgM5ykjY">wave function</a> of the ground state of a <a href="/facts/Particle_in_a_box/rezR97ky">particle in a one-dimensional box</a> is a half-period <a href="/facts/Sine_wave/mD1CB09M">sine wave</a>, which goes to zero at the two edges of the well. The energy of the particle is given by h 2 n 2 8 m L 2 {\textstyle {\frac {h^{2}n^{2}}{8mL^{2}}}} , where <i>h</i> is the <a href="/facts/Planck_constant/qGHAe3qq">Planck constant</a>, <i>m</i> is the mass of the particle, <i>n</i> is the energy state (<i>n</i> = 1 corresponds to the ground-state energy), and <i>L</i> is the width of the well.</li> <li>The wave function of the ground state of a hydrogen atom is a spherically symmetric distribution centred on the <a href="/facts/Atomic_nucleus/orpIm5Xl">nucleus</a>, which is largest at the center and reduces <a href="/facts/Exponential_distribution/yY3EdV0u">exponentially</a> at larger distances. The <a href="/facts/Electron/L0KBo5cW">electron</a> is most likely to be found at a distance from the nucleus equal to the <a href="/facts/Bohr_radius/vvP1lghl">Bohr radius</a>. This function is known as the 1s <a href="/facts/Atomic_orbital/tKhgl6mL">atomic orbital</a>. For hydrogen (H), an electron in the ground state has energy −13.6 eV, relative to the <a href="/facts/Ionization_energy/lezUajfA">ionization threshold</a>. In other words, 13.6 eV is the energy input required for the electron to no longer be <a href="/facts/Bound_state/NndW5JQl">bound</a> to the atom.</li> <li>The exact definition of one <a href="/facts/Second/jtFxnJXx">second</a> of <a href="/facts/Time/HceJKUGB">time</a> since 1997 has been the duration of 9192631770 periods of the radiation corresponding to the transition between the two <a href="/facts/Hyperfine_structure/aqj0dMut"> hyperfine</a> levels of the ground state of the <a href="/facts/Caesium/dXKN1S5k">caesium</a>-133 atom at rest at a temperature of 0 K.<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a></li></ul> <h2 id="notes">Notes</h2> <h2 id="bibliography">Bibliography</h2> <ul><li><a href="/facts/Richard_Feynman/32vGFrtB">Feynman, Richard</a>; Leighton, Robert; Sands, Matthew (1965). <a href="https://feynmanlectures.caltech.edu/III_toc.html">"see section 2-5 for energy levels, 19 for the hydrogen atom"</a>. <i>The Feynman Lectures on Physics</i>. Vol. 3.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>See, 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 as Feynman, 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. <a href="https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf" target="_blank">https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>See, 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 as Feynman, 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. <a href="https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf" target="_blank">https://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>i.e. ⟨ ψ 1 | ψ 2 ⟩ = δ i j {\displaystyle \left\langle \psi _{1}|\psi _{2}\right\rangle =\delta _{ij}} <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>"Unit of time (second)". SI Brochure. International Bureau of Weights and Measures. Retrieved 2013-12-22. <a href="http://www.bipm.org/en/si/si_brochure/chapter2/2-1/second.html" target="_blank">http://www.bipm.org/en/si/si_brochure/chapter2/2-1/second.html</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> </ol>