Ground state

Derivation

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

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

where V(x) is the potential.

With integration by parts:

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

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

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

Assuming ${\displaystyle \psi (x)\approx -cx}$ around ${\displaystyle x=0}$, one may write $${\displaystyle \psi '(x)=N{\begin{cases}|\psi (x)|,&|x|>\varepsilon ,\\c\varepsilon ,&|x|\leq \varepsilon ,\end{cases}}}$$ where ${\displaystyle N={\frac {1}{\sqrt {1+{\frac {4}{3}}|c|^{2}\varepsilon ^{3}}}}}$ 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 average kinetic energy is lowered by ${\displaystyle O(\varepsilon )}$ by the deformation to ψ'.

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

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

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

We can therefore remove all nodes and reduce the energy by ${\displaystyle O(\varepsilon )}$, which implies that ${\displaystyle \psi }$ 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 quantum states with the energy eigenvalue of the ground state (let's name it ${\displaystyle E_{g}}$) and the same spin state and therefore would only differ in their position-space wave functions.

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

Now let ${\displaystyle x_{0}}$ be some random point (where both wave functions are defined) and set: $${\displaystyle c_{1}={\frac {\psi _{2}(x_{0},0)}{a}}}$$ and $${\displaystyle c_{2}={\frac {-\psi _{1}(x_{0},0)}{a}}}$$ with $${\displaystyle a={\sqrt {|\psi _{1}(x_{0},0)|^{2}+|\psi _{2}(x_{0},0)|^{2}}}>0}$$ (according to the premise no nodes).

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

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

But ${\displaystyle \left\langle \psi _{3}|\psi _{3}\right\rangle =|c_{1}|^{2}+|c_{2}|^{2}=1}$ i.e., ${\displaystyle x_{0}}$ 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 ${\displaystyle \left|\uparrow \right\rangle }$ 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.