Hamiltonian (quantum mechanics)

By analogy with classical mechanics, the Hamiltonian is commonly expressed as the sum of operators corresponding to the kinetic and potential energies of a system in the form

$${\displaystyle {\hat {H}}={\hat {T}}+{\hat {V}},}$$

where

$${\displaystyle {\hat {V}}=V=V(\mathbf {r} ,t),}$$

is the potential energy operator and

$${\displaystyle {\hat {T}}={\frac {\mathbf {\hat {p}} \cdot \mathbf {\hat {p}} }{2m}}={\frac {{\hat {p}}^{2}}{2m}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2},}$$

is the kinetic energy operator in which ${\displaystyle m}$ is the mass of the particle, the dot denotes the dot product of vectors, and

$${\displaystyle {\hat {p}}=-i\hbar \nabla ,}$$

is the momentum operator where a ${\displaystyle \nabla }$ is the del operator. The dot product of ${\displaystyle \nabla }$ with itself is the Laplacian ${\displaystyle \nabla ^{2}}$. In three dimensions using Cartesian coordinates the Laplace operator is

$${\displaystyle \nabla ^{2}={\frac {\partial ^{2}}{{\partial x}^{2}}}+{\frac {\partial ^{2}}{{\partial y}^{2}}}+{\frac {\partial ^{2}}{{\partial z}^{2}}}}$$

Although this is not the technical definition of the Hamiltonian in classical mechanics, it is the form it most commonly takes. Combining these yields the familiar form used in the Schrödinger equation:

$${\displaystyle {\begin{aligned}{\hat {H}}&={\hat {T}}+{\hat {V}}\\[6pt]&={\frac {\mathbf {\hat {p}} \cdot \mathbf {\hat {p}} }{2m}}+V(\mathbf {r} ,t)\\[6pt]&=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(\mathbf {r} ,t)\end{aligned}}}$$

which allows one to apply the Hamiltonian to systems described by a wave function ${\displaystyle \Psi (\mathbf {r} ,t)}$. This is the approach commonly taken in introductory treatments of quantum mechanics, using the formalism of Schrödinger's wave mechanics.

One can also make substitutions to certain variables to fit specific cases, such as some involving electromagnetic fields.

The formalism can be extended to ${\displaystyle N}$ particles:

$${\displaystyle {\hat {H}}=\sum _{n=1}^{N}{\hat {T}}_{n}+{\hat {V}}}$$

where

$${\displaystyle {\hat {V}}=V(\mathbf {r} _{1},\mathbf {r} _{2},\ldots ,\mathbf {r} _{N},t),}$$

is the potential energy function, now a function of the spatial configuration of the system and time (a particular set of spatial positions at some instant of time defines a configuration) and

$${\displaystyle {\hat {T}}_{n}={\frac {\mathbf {p} _{n}\cdot \mathbf {p} _{n}}{2m_{n}}}=-{\frac {\hbar ^{2}}{2m_{n}}}\nabla _{n}^{2}}$$

is the kinetic energy operator of particle ${\displaystyle n}$, ${\displaystyle \nabla _{n}}$ is the gradient for particle ${\displaystyle n}$, and ${\displaystyle \nabla _{n}^{2}}$ is the Laplacian for particle n:

$${\displaystyle \nabla _{n}^{2}={\frac {\partial ^{2}}{\partial x_{n}^{2}}}+{\frac {\partial ^{2}}{\partial y_{n}^{2}}}+{\frac {\partial ^{2}}{\partial z_{n}^{2}}},}$$

Combining these yields the Schrödinger Hamiltonian for the ${\displaystyle N}$-particle case:

$${\displaystyle {\begin{aligned}{\hat {H}}&=\sum _{n=1}^{N}{\hat {T}}_{n}+{\hat {V}}\\[6pt]&=\sum _{n=1}^{N}{\frac {\mathbf {\hat {p}} _{n}\cdot \mathbf {\hat {p}} _{n}}{2m_{n}}}+V(\mathbf {r} _{1},\mathbf {r} _{2},\ldots ,\mathbf {r} _{N},t)\\[6pt]&=-{\frac {\hbar ^{2}}{2}}\sum _{n=1}^{N}{\frac {1}{m_{n}}}\nabla _{n}^{2}+V(\mathbf {r} _{1},\mathbf {r} _{2},\ldots ,\mathbf {r} _{N},t)\end{aligned}}}$$

However, complications can arise in the many-body problem. Since the potential energy depends on the spatial arrangement of the particles, the kinetic energy will also depend on the spatial configuration to conserve energy. The motion due to any one particle will vary due to the motion of all the other particles in the system. For this reason cross terms for kinetic energy may appear in the Hamiltonian; a mix of the gradients for two particles:

$${\displaystyle -{\frac {\hbar ^{2}}{2M}}\nabla _{i}\cdot \nabla _{j}}$$

where ${\displaystyle M}$ denotes the mass of the collection of particles resulting in this extra kinetic energy. Terms of this form are known as mass polarization terms, and appear in the Hamiltonian of many electron atoms (see below).

For ${\displaystyle N}$ interacting particles, i.e. particles which interact mutually and constitute a many-body situation, the potential energy function ${\displaystyle V}$ is not simply a sum of the separate potentials (and certainly not a product, as this is dimensionally incorrect). The potential energy function can only be written as above: a function of all the spatial positions of each particle.

For non-interacting particles, i.e. particles which do not interact mutually and move independently, the potential of the system is the sum of the separate potential energy for each particle,[1] that is

$${\displaystyle V=\sum _{i=1}^{N}V(\mathbf {r} _{i},t)=V(\mathbf {r} _{1},t)+V(\mathbf {r} _{2},t)+\cdots +V(\mathbf {r} _{N},t)}$$

The general form of the Hamiltonian in this case is:

$${\displaystyle {\begin{aligned}{\hat {H}}&=-{\frac {\hbar ^{2}}{2}}\sum _{i=1}^{N}{\frac {1}{m_{i}}}\nabla _{i}^{2}+\sum _{i=1}^{N}V_{i}\\[6pt]&=\sum _{i=1}^{N}\left(-{\frac {\hbar ^{2}}{2m_{i}}}\nabla _{i}^{2}+V_{i}\right)\\[6pt]&=\sum _{i=1}^{N}{\hat {H}}_{i}\end{aligned}}}$$

where the sum is taken over all particles and their corresponding potentials; the result is that the Hamiltonian of the system is the sum of the separate Hamiltonians for each particle. This is an idealized situation—in practice the particles are almost always influenced by some potential, and there are many-body interactions. One illustrative example of a two-body interaction where this form would not apply is for electrostatic potentials due to charged particles, because they interact with each other by Coulomb interaction (electrostatic force), as shown below.

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

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}}$$

and in higher dimensions:

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}}$$

For a particle in a region of constant potential ${\displaystyle V=V_{0}}$ (no dependence on space or time), in one dimension, the Hamiltonian is:

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V_{0}}$$

in three dimensions

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V_{0}}$$

This applies to the elementary "particle in a box" problem, and step potentials.

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

$${\displaystyle V={\frac {k}{2}}x^{2}={\frac {m\omega ^{2}}{2}}x^{2}}$$

where the angular frequency ${\displaystyle \omega }$, effective spring constant ${\displaystyle k}$, and mass ${\displaystyle m}$ of the oscillator satisfy:

$${\displaystyle \omega ^{2}={\frac {k}{m}}}$$

so the Hamiltonian is:

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {m\omega ^{2}}{2}}x^{2}}$$

For three dimensions, this becomes

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+{\frac {m\omega ^{2}}{2}}r^{2}}$$

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

$${\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} =|\mathbf {r} |^{2}=x^{2}+y^{2}+z^{2}}$$

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

$${\displaystyle {\begin{aligned}{\hat {H}}&=-{\frac {\hbar ^{2}}{2m}}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)+{\frac {m\omega ^{2}}{2}}(x^{2}+y^{2}+z^{2})\\[6pt]&=\left(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {m\omega ^{2}}{2}}x^{2}\right)+\left(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {m\omega ^{2}}{2}}y^{2}\right)+\left(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial z^{2}}}+{\frac {m\omega ^{2}}{2}}z^{2}\right)\end{aligned}}}$$

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, say due to double or triple chemical bonds), the Hamiltonian is:

$${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2I_{xx}}}{\hat {J}}_{x}^{2}-{\frac {\hbar ^{2}}{2I_{yy}}}{\hat {J}}_{y}^{2}-{\frac {\hbar ^{2}}{2I_{zz}}}{\hat {J}}_{z}^{2}}$$

where ${\displaystyle I_{xx}}$, ${\displaystyle I_{yy}}$, and ${\displaystyle I_{zz}}$ are the moment of inertia components (technically the diagonal elements of the moment of inertia tensor), and ${\displaystyle {\hat {J}}_{x}}$, ${\displaystyle {\hat {J}}_{y}}$, and ${\displaystyle {\hat {J}}_{z}}$ are the total angular momentum operators (components), about the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ axes respectively.

The Coulomb potential energy for two point charges ${\displaystyle q_{1}}$ and ${\displaystyle q_{2}}$ (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):

$${\displaystyle V={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}|\mathbf {r} |}}}$$

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 ${\displaystyle N}$ charges, the potential energy of charge ${\displaystyle q_{j}}$ due to all other charges is (see also Electrostatic potential energy stored in a configuration of discrete point charges):[3]

$${\displaystyle V_{j}={\frac {1}{2}}\sum _{i\neq j}q_{i}\phi (\mathbf {r} _{i})={\frac {1}{8\pi \varepsilon _{0}}}\sum _{i\neq j}{\frac {q_{i}q_{j}}{|\mathbf {r} _{i}-\mathbf {r} _{j}|}}}$$

where ${\displaystyle \phi (\mathbf {r} _{i})}$ is the electrostatic potential of charge ${\displaystyle q_{j}}$ at ${\displaystyle \mathbf {r} _{i}}$. The total potential of the system is then the sum over ${\displaystyle j}$:

$${\displaystyle V={\frac {1}{8\pi \varepsilon _{0}}}\sum _{j=1}^{N}\sum _{i\neq j}{\frac {q_{i}q_{j}}{|\mathbf {r} _{i}-\mathbf {r} _{j}|}}}$$

so the Hamiltonian is:

$${\displaystyle {\begin{aligned}{\hat {H}}&=-{\frac {\hbar ^{2}}{2}}\sum _{j=1}^{N}{\frac {1}{m_{j}}}\nabla _{j}^{2}+{\frac {1}{8\pi \varepsilon _{0}}}\sum _{j=1}^{N}\sum _{i\neq j}{\frac {q_{i}q_{j}}{|\mathbf {r} _{i}-\mathbf {r} _{j}|}}\\&=\sum _{j=1}^{N}\left(-{\frac {\hbar ^{2}}{2m_{j}}}\nabla _{j}^{2}+{\frac {1}{8\pi \varepsilon _{0}}}\sum _{i\neq j}{\frac {q_{i}q_{j}}{|\mathbf {r} _{i}-\mathbf {r} _{j}|}}\right)\\\end{aligned}}}$$

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

$${\displaystyle V=-\mathbf {\hat {d}} \cdot \mathbf {E} }$$

the dipole moment itself is the operator

$${\displaystyle \mathbf {\hat {d}} =q\mathbf {\hat {r}} }$$

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:

$${\displaystyle {\hat {H}}=-\mathbf {\hat {d}} \cdot \mathbf {E} =-q\mathbf {\hat {r}} \cdot \mathbf {E} }$$

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

$${\displaystyle V=-{\boldsymbol {\mu }}\cdot \mathbf {B} }$$

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:

$${\displaystyle {\hat {H}}=-{\boldsymbol {\mu }}\cdot \mathbf {B} }$$

For a spin-1⁄2 particle, the corresponding spin magnetic moment is:[4]

$${\displaystyle {\boldsymbol {\mu }}_{S}={\frac {g_{s}e}{2m}}\mathbf {S} }$$

where ${\displaystyle g_{s}}$ is the "spin g-factor" (not to be confused with the gyromagnetic ratio), ${\displaystyle e}$ is the electron charge, ${\displaystyle \mathbf {S} }$ is the spin operator vector, whose components are the Pauli matrices, hence

$${\displaystyle {\hat {H}}={\frac {g_{s}e}{2m}}\mathbf {S} \cdot \mathbf {B} }$$

For a particle with mass ${\displaystyle m}$ and charge ${\displaystyle q}$ in an electromagnetic field, described by the scalar potential ${\displaystyle \phi }$ and vector potential ${\displaystyle \mathbf {A} }$, there are two parts to the Hamiltonian to substitute for.[1] The canonical momentum operator ${\displaystyle \mathbf {\hat {\Pi }} }$, which includes a contribution from the ${\displaystyle \mathbf {A} }$ field and fulfils the canonical commutation relation, must be quantized;

$${\displaystyle \mathbf {\hat {\Pi }} =\mathbf {\hat {P}} +q\mathbf {A} ,}$$

where ${\displaystyle \mathbf {\hat {P}} }$ is the kinetic momentum operator. The quantization prescription reads

$${\displaystyle \mathbf {\hat {\Pi }} =-i\hbar \nabla ,}$$

so the corresponding kinetic energy operator is

$${\displaystyle {\hat {T}}={\frac {\mathbf {\hat {P}} \cdot \mathbf {\hat {P}} }{2m}}={\frac {1}{2m}}\left(\mathbf {\hat {\Pi }} -q\mathbf {A} \right)^{2}}$$

and the potential energy, which is due to the ${\displaystyle \phi }$ field, is given by

$${\displaystyle {\hat {V}}=q\phi .}$$

Casting all of these into the Hamiltonian gives

$${\displaystyle {\hat {H}}={\frac {1}{2m}}\left(-i\hbar \nabla -q\mathbf {A} \right)^{2}+q\phi .}$$