Editing Hamiltonian (quantum mechanics) (section)

==Expressions for the Hamiltonian==

Following are expressions for the Hamiltonian in a number of situations.<ref>{{cite book |title=Quanta: A Handbook of Concepts |first=P. W. |last=Atkins |publisher=Oxford University Press |year=1974 |isbn=0-19-855493-1 }}</ref> Typical ways to classify the expressions are the number of particles, number of dimensions, and the nature of the potential energy function—importantly space and time dependence. Masses are denoted by <math>m</math>, and charges by <math>q</math>.

===Free particle===

The particle is not bound by any potential energy, so the potential is zero and this Hamiltonian is the simplest. For one dimension:

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} </math>

and in higher dimensions:

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 </math>

===Constant-potential well===

For a particle in a region of constant potential <math>V = V_0</math> (no dependence on space or time), in one dimension, the Hamiltonian is:

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V_0 </math>

in three dimensions

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V_0 </math>

This applies to the elementary "[[particle in a box]]" problem, and [[step potential]]s.

===Simple harmonic oscillator===

For a [[simple harmonic oscillator]] in one dimension, the potential varies with position (but not time), according to:

<math display="block">V = \frac{k}{2}x^2 = \frac{m\omega^2}{2}x^2  </math>

where the [[angular frequency]] <math>\omega</math>, effective [[spring constant]] <math>k</math>, and mass <math>m</math> of the oscillator satisfy:

<math display="block">\omega^2 = \frac{k}{m}</math>

so the Hamiltonian is:

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{m\omega^2}{2}x^2 </math>

For three dimensions, this becomes

<math display="block">\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + \frac{m\omega^2}{2} r^2 </math>

where the three-dimensional position vector <math>\mathbf{r}</math> using Cartesian coordinates is <math>(x, y, z)</math>, its magnitude is

<math display="block">r^2 = \mathbf{r}\cdot\mathbf{r} = |\mathbf{r}|^2 = x^2+y^2+z^2 </math>

Writing the Hamiltonian out in full shows it is simply the sum of the one-dimensional Hamiltonians in each direction:

<math display="block">\begin{align}
\hat{H} & = -\frac{\hbar^2}{2m}\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) + \frac{m\omega^2}{2} \left(x^2 + y^2 + z^2\right) \\[6pt]
& = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{m\omega^2}{2}x^2\right) + \left(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial y^2} + \frac{m\omega^2}{2}y^2 \right ) + \left(- \frac{\hbar^2}{2m}\frac{\partial^2}{\partial z^2} +\frac{m\omega^2}{2}z^2 \right)
\end{align}</math>

===Rigid rotor===

For a [[rigid rotor]]—i.e., system of particles which can rotate freely about any axes, not bound in any potential (such as free molecules with negligible vibrational [[Degrees of freedom (physics and chemistry)|degrees of freedom]], say due to [[double bond|double]] or [[triple bond|triple]] [[chemical bond]]s), the Hamiltonian is:

<math display="block"> \hat{H} = -\frac{\hbar^2}{2I_{xx}}\hat{J}_x^2 -\frac{\hbar^2}{2I_{yy}}\hat{J}_y^2 -\frac{\hbar^2}{2I_{zz}}\hat{J}_z^2 </math>

where <math>I_{xx}</math>, <math>I_{yy}</math>, and <math>I_{zz}</math> are the [[moment of inertia]] components (technically the diagonal elements of the [[Moment of inertia#Moment of inertia tensor|moment of inertia tensor]]), and {{nowrap|<math> \hat{J}_x</math>,}} {{nowrap|<math> \hat{J}_y</math>,}} and <math> \hat{J}_z</math> are the total [[angular momentum]] operators (components), about the <math>x</math>, <math>y</math>, and <math>z</math> axes respectively.

===Electrostatic (Coulomb) potential===

The [[Coulomb potential energy]] for two point charges <math>q_1</math> and <math>q_2</math> (i.e., those that have no spatial extent independently), in three dimensions, is (in [[SI units]]—rather than [[Gaussian units]] which are frequently used in [[electromagnetism]]):

<math display="block">V = \frac{q_1q_2}{4\pi\varepsilon_0 |\mathbf{r}|}</math>

However, this is only the potential for one point charge due to another. If there are many charged particles, each charge has a potential energy due to every other point charge (except itself). For <math>N</math> charges, the potential energy of charge <math>q_j</math> due to all other charges is (see also [[Electric potential energy#Electrostatic potential energy stored in a system of point charges|Electrostatic potential energy stored in a configuration of discrete point charges]]):<ref>{{cite book |title=Electromagnetism |url=https://archive.org/details/electromagnetism0000gran |url-access=registration |edition=2nd |first1=I. S. |last1=Grant |first2=W. R. |last2=Phillips |series=Manchester Physics Series |year=2008 |isbn=978-0-471-92712-9 }}</ref>

<math display="block">V_j = \frac{1}{2}\sum_{i\neq j} q_i \phi(\mathbf{r}_i)=\frac{1}{8\pi\varepsilon_0}\sum_{i\neq j} \frac{q_iq_j}{|\mathbf{r}_i-\mathbf{r}_j|}</math>

where <math>\phi(\mathbf{r}_i)</math> is the electrostatic potential of charge <math>q_j</math> at <math>\mathbf{r}_i</math>. The total potential of the system is then the sum over <math>j</math>:

<math display="block">V = \frac{1}{8\pi\varepsilon_0}\sum_{j=1}^N\sum_{i\neq j} \frac{q_iq_j}{|\mathbf{r}_i-\mathbf{r}_j|}</math>

so the Hamiltonian is:

<math display="block">\begin{align}
\hat{H} & = -\frac{\hbar^2}{2}\sum_{j=1}^N\frac{1}{m_j}\nabla_j^2 + \frac{1}{8\pi\varepsilon_0}\sum_{j=1}^N\sum_{i\neq j} \frac{q_iq_j}{|\mathbf{r}_i-\mathbf{r}_j|} \\
 & = \sum_{j=1}^N \left ( -\frac{\hbar^2}{2m_j}\nabla_j^2 + \frac{1}{8\pi\varepsilon_0}\sum_{i\neq j} \frac{q_iq_j}{|\mathbf{r}_i-\mathbf{r}_j|}\right) \\
\end{align}</math>

===Electric dipole in an electric field===

For an [[electric dipole moment]] <math>\mathbf{d}</math> constituting charges of magnitude <math>q</math>, in a uniform, [[electrostatic field]] (time-independent) <math>\mathbf{E}</math>, positioned in one place, the potential is:

<math display="block">V = -\mathbf{\hat{d}}\cdot\mathbf{E} </math>

the dipole moment itself is the operator

<math display="block">\mathbf{\hat{d}} = q\mathbf{\hat{r}} </math>

Since the particle is stationary, there is no translational kinetic energy of the dipole, so the Hamiltonian of the dipole is just the potential energy:

<math display="block">\hat{H} = -\mathbf{\hat{d}}\cdot\mathbf{E} = -q\mathbf{\hat{r}}\cdot\mathbf{E}</math>

===Magnetic dipole in a magnetic field===

For a magnetic dipole moment <math>\boldsymbol{\mu}</math> in a uniform, magnetostatic field (time-independent) <math>\mathbf{B}</math>, positioned in one place, the potential is:

<math display="block">V = -\boldsymbol{\mu}\cdot\mathbf{B} </math>

Since the particle is stationary, there is no translational kinetic energy of the dipole, so the Hamiltonian of the dipole is just the potential energy:

<math display="block">\hat{H} = -\boldsymbol{\mu}\cdot\mathbf{B} </math>

For a [[spin-1/2|spin-{{frac|1|2}}]] particle, the corresponding spin magnetic moment is:<ref>{{cite book |title=Physics of Atoms and Molecules |first1=B. H. |last1=Bransden |first2=C. J. |last2=Joachain |publisher=Longman |year=1983 |isbn=0-582-44401-2 }}</ref>

<math display="block">\boldsymbol{\mu}_S = \frac{g_s e}{2m} \mathbf{S} </math>

where <math>g_s</math> is the "spin [[g-factor (physics)|g-factor]]" (not to be confused with the [[gyromagnetic ratio]]), <math>e</math> is the electron charge, <math>\mathbf{S}</math> is the [[Spin (physics)#Pauli matrices and spin operators|spin operator]] vector, whose components are the [[Pauli matrices]], hence

<math display="block">\hat{H} = \frac{g_s e}{2m} \mathbf{S} \cdot\mathbf{B} </math>

===Charged particle in an electromagnetic field===

For a particle with mass <math>m</math> and charge <math>q</math> in an electromagnetic field, described by the [[scalar potential]] <math>\phi</math> and [[vector potential]] <math>\mathbf{A}</math>, there are two parts to the Hamiltonian to substitute for.<ref name="QuantumPhysics" /> The canonical momentum operator <math>\mathbf{\hat{p}}</math>, which includes a contribution from the <math>\mathbf{A}</math> field and fulfils the [[canonical commutation relation]], must be quantized;

<math display="block">\mathbf{\hat{p}} = m\dot{\mathbf{r}} + q\mathbf{A} ,</math>

where <math>m\dot{\mathbf{r}}</math> is the [[kinetic momentum]]. The quantization prescription reads

<math display="block">\mathbf{\hat{p}} = -i\hbar\nabla ,</math>

so the corresponding kinetic energy operator is

<math display="block">\hat{T} = \frac{1}{2} m\dot{\mathbf{r}}\cdot\dot{\mathbf{r}} = \frac{1}{2m} \left ( \mathbf{\hat{p}} - q\mathbf{A} \right)^2 </math>

and the potential energy, which is due to the <math>\phi</math> field, is given by

<math display="block">\hat{V} = q\phi .</math>

Casting all of these into the Hamiltonian gives

<math display="block">\hat{H} = \frac{1}{2m} \left ( -i\hbar\nabla - q\mathbf{A} \right)^2 + q\phi .</math>