Editing Hamiltonian (quantum mechanics) (section)

===One particle===

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

<math display="block"> \hat{H} = \hat{T} + \hat{V}, </math>

where
<math display="block"> \hat{V} = V = V(\mathbf{r},t) ,</math>
is the [[potential energy]] operator and
<math display="block">\hat{T} = \frac{\mathbf{\hat{p}}\cdot\mathbf{\hat{p}}}{2m} = \frac{\hat{p}^2}{2m} = -\frac{\hbar^2}{2m}\nabla^2,</math>
is the [[kinetic energy]] operator in which <math>m</math> is the [[mass]] of the particle, the dot denotes the [[dot product]] of vectors, and
<math display="block"> \hat{p} = -i\hbar\nabla ,</math>
is the [[momentum operator]] where a <math>\nabla</math> is the [[del]] [[operator (mathematics)|operator]]. The [[dot product]] of <math>\nabla</math> with itself is the [[Laplacian]] <math>\nabla^2</math>. In three dimensions using [[Cartesian coordinates]] the Laplace operator is
<math display="block">\nabla^2 = \frac{\partial^2}{ {\partial x}^2} + \frac{\partial^2}{ {\partial y}^2} + \frac{\partial^2}{ {\partial z}^2}</math>

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

<math display="block">\begin{align}
\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{align}</math>

which allows one to apply the Hamiltonian to systems described by a [[wave function]] <math>\Psi(\mathbf{r}, t)</math>. 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.

==== Expectation value ====
It can be shown that the expectation value of the Hamiltonian which gives the energy expectation value will always be greater than or equal to the minimum potential of the system.

Consider computing the expectation value of kinetic energy:

<math display="block">\begin{align}
T &= -\frac{\hbar^2}{2m} \int_{-\infty}^{+\infty} \psi^* \frac{d^2\psi}{dx^2} \, dx \\[1ex]
&=-\frac{\hbar^2}{2m} \left( {\left[ \psi'(x) \psi^*(x) \right]}_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \frac{d\psi}{dx} \frac{d\psi^*}{dx} \, dx \right) \\[1ex]
&= \frac{\hbar^2}{2m} \int_{-\infty}^{+\infty} \left|\frac{d\psi}{dx}  \right|^2 \, dx  \geq 0
\end{align}</math>

Hence the expectation value of kinetic energy is always non-negative. This result can be used to calculate the expectation value of the total energy which is given for a normalized wavefunction as:

<math display="block">E = T + \langle V(x) \rangle = T + \int_{-\infty}^{+\infty} V(x) |\psi(x)|^2 \, dx \geq  V_{\text{min}}(x) \int_{-\infty}^{+\infty} |\psi(x)|^2 \, dx  \geq V_{\text{min}}(x)  </math>

which complete the proof. Similarly, the condition can be generalized to any higher dimensions using [[divergence theorem]].