Ground state

 ( Quantum States ) 

The ground state of a 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 vacuum.

If more than one ground state exists, they are said to be 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 with the 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.

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Absence of nodes in one dimension In one dimension, the ground state of the Schrödinger equation can be proven to have no nodes. See, for example, Published as

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Derivation Consider the average energy of a state with a node at ; i.e., . The average energy in this state would be

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

where is the potential.

With integration by parts:

$$\backslash int\_a^b\; \backslash psi^*\; \backslash frac\{d^2\backslash psi\}\{dx^2\}\; dx\; =\; \backslash left\_a^b\; -\; \backslash int\_a^b\; \backslash frac\{d\backslash psi^*\}\{dx\}\backslash frac\{d\backslash psi\}\{dx\}\; dx\; =\; \backslash left\_a^b\; -\; \backslash int\_a^b\; \backslash left|\backslash frac\{d\backslash psi\}\{dx\}\backslash right|^2\; dx$$

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

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

Assuming $\backslash psi(x)\; \backslash approx\; -cx$ around $x\; =\; 0$, one may write $$$$

 \psi'(x) = N \begin{cases}
   |\psi(x)|,    & |x| >   \varepsilon, \\
   c\varepsilon, & |x| \le \varepsilon,
 \end{cases}
     

where $N\; =\; \backslash frac\{1\}\{\backslash sqrt\{1\; +\; \backslash frac\{4\}\{3\}\; |c|^2\backslash varepsilon^3\}\}$ is the norm.

Note that the kinetic-energy densities hold everywhere because of the normalization. More significantly, the average kinetic energy is lowered by $O(\backslash varepsilon)$ by the deformation to .

Now, consider the potential energy. For definiteness, let us choose $V(x)\; \backslash ge\; 0$. Then it is clear that, outside the interval $x\; \backslash in\; -\backslash varepsilon,$, the potential energy density is smaller for the because there.

On the other hand, in the interval $x\; \backslash in\; -\backslash varepsilon,$ we have $$$$

 {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,
     

which holds to order $\backslash varepsilon^3$.

However, the contribution to the potential energy from this region for the state with a node is $$$$

 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,
     

lower, but still of the same lower order $O(\backslash varepsilon^3)$ as for the deformed state , and subdominant to the lowering of the average kinetic energy. Therefore, the potential energy is unchanged up to order $\backslash varepsilon^2$, if we deform the state $\backslash psi$ with a node into a state without a node, and the change can be ignored.

We can therefore remove all nodes and reduce the energy by $O(\backslash varepsilon)$, which implies that 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.)

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Implication As the ground state has no nodes it is spatially non-degenerate, i.e. there are no two Stationary state with the energy eigenvalue of the ground state (let's name it $E\_g$) and the same spin state and therefore would only differ in their position-space .

The reasoning goes by contradiction: For if the ground state would be degenerate then there would be two orthonormali.e. $\backslash left\backslash lang\; \backslash psi\_1|\backslash psi\_2\backslash right\backslash rang\; =\; \backslash delta\_\{ij\}$ stationary states $\backslash left|\backslash psi\_1\backslash right\backslash rang$ and $\backslash left|\backslash psi\_2\backslash right\backslash rang$ — later on represented by their complex-valued position-space wave functions $\backslash psi\_1(x,t)=\backslash psi\_1(x,0)\backslash cdot\; e^\{-iE\_g\; t/\backslash hbar\}$ and $\backslash psi\_2(x,t)=\backslash psi\_2(x,0)\backslash cdot\; e^\{-iE\_g\; t/\backslash hbar\}$ — and any superposition $\backslash left|\backslash psi\_3\backslash right\backslash rang\; :=\; c\_1\backslash left|\backslash psi\_1\backslash right\backslash rang\; +\; c\_2\backslash left|\backslash psi\_2\backslash right\backslash rang$ 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: $$c\_1=\backslash frac\{\backslash psi\_2(x\_0,0)\}\{a\}$$ and $$c\_2=\backslash frac\{-\backslash psi\_1(x\_0,0)\}\{a\}$$ with $$a=\backslash sqrt\{|\backslash psi\_1(x\_0,0)|^2+|\backslash psi\_2(x\_0,0)|^2\}\; >\; 0$$ (according to the premise no nodes).

Therefore, the position-space wave function of $\backslash left|\backslash psi\_3\backslash right\backslash rang$ is $$\backslash psi\_3(x,t)=c\_1\backslash psi\_1(x,t)+c\_2\backslash psi\_2(x,t)\; =\; \backslash frac\{1\}\{a\}\backslash left(\backslash psi\_2(x\_0,0)\backslash cdot\backslash psi\_1(x,0)\; -\; \backslash psi\_1(x\_0,0)\backslash cdot\backslash psi\_2(x,0)\; \backslash right)\backslash cdot\; e^\{-iE\_g\; t/\backslash hbar\}.$$

Hence $$\backslash psi\_3(x\_0,t)=\backslash frac\{1\}\{a\}\backslash left(\backslash psi\_2(x\_0,0)\backslash cdot\backslash psi\_1(x\_0,0)\; -\; \backslash psi\_1(x\_0,0)\backslash cdot\backslash psi\_2(x\_0,0)\; \backslash right)\backslash cdot\; e^\{-iE\_g\; t/\backslash hbar\}\; =\; 0$$ for all $t$.

But $\backslash left\backslash lang\; \backslash psi\_3|\backslash psi\_3\backslash right\backslash rang\; =\; |c\_1|^2+|c\_2|^2=1$ i.e., $x\_0$ 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 $\backslash left|\backslash uparrow\backslash right\backslash rang$ and $\backslash left|\backslash downarrow\backslash right\backslash rang$ 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.

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Examples

• The wave function of the ground state of a 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 $\backslash frac\{h^2\; n^2\}\{8\; m\; L^2\}$, 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, which is largest at the center and reduces 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 , relative to the ionization threshold. In other words, 13.6 eV is the energy input required for the electron to no longer be Bound state to the atom.
• The exact definition of one second of time since 1997 has been the duration of periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom at rest at a temperature of 0 K.

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Notes

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Bibliography

Categories: Quantum States

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