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quantum mechanics: see quantum theory.

The Columbia Electronic Encyclopedia Copyright © 2004.

Licensed from Columbia University Press

Licensed from Columbia University Press

In quantum mechanics, the Hamiltonian H is the observable corresponding to the total energy of the system. As with all observables, the spectrum of the Hamiltonian is the set of possible outcomes when one measures the total energy of a system. Like any other self-adjoint operator, the spectrum of the Hamiltonian can be decomposed, via its spectral measures, into pure point, absolutely continuous, and singular parts. The pure point spectrum can be associated to eigenvectors, which in turn are the bound states of the system. The absolutely continuous spectrum corresponds to the free states. The singular spectrum, interestingly enough, comprises physically impossible outcomes. For example, consider the finite potential well, which admits bound states with discrete negative energies and free states with continuous positive energies.

- $H\; left|\; psi\; (t)\; rightrangle\; =\; mathrm\{i\}\; hbar\; \{partialoverpartial\; t\}\; left|\; psi\; (t)\; rightrangle$.

where $hbar$ is the reduced Planck constant . This equation is known as the Schrödinger equation. (It takes the same form as the Hamilton-Jacobi equation, which is one of the reasons H is also called the Hamiltonian.) Given the state at some initial time (t = 0), we can integrate it to obtain the state at any subsequent time. In particular, if H is independent of time, then

- $left|\; psi\; (t)\; rightrangle\; =\; expleft(-\{mathrm\{i\}Ht\; over\; hbar\}right)\; left|\; psi\; (0)\; rightrangle$.

Note: In introductory physics literature, the following is often taken as an assumption:

- The eigenkets (eigenvectors) of H, denoted $left|\; a\; rightrang$ (using Dirac bra-ket notation), provide an orthonormal basis for the Hilbert space. The spectrum of allowed energy levels of the system is given by the set of eigenvalues, denoted {E
_{a}}, solving the equation:

- $H\; left|\; a\; rightrangle\; =\; E\_a\; left|\; a\; rightrangle$.

- Since H is a Hermitian operator, the energy is always a real number.

From a mathematically rigorous point of view, care must be taken with the above assumption. Operators on infinite-dimensional Hilbert spaces need not have eigenvalues (the set of eigenvalues does not necessarily coincide with the spectrum of an operator). However, all routine quantum mechanical calculations can be done using the physical formulation.

Similarly, the exponential operator on the right hand side of the Schrödinger equation is sometimes defined by the power series. One might notice that taking polynomials of unbounded and not everywhere defined operators may not make mathematical sense, much less power series. Rigorously, to take functions of unbounded operators, a functional calculus is required. In the case of the exponential function, the continuous, or just the holomorphic functional calculus suffices. We note again, however, that for common calculations the physicist's formulation is quite sufficient.

By the *-homomorphism property of the functional calculus, the operator

- $U\; =\; expleft(-\{mathrm\{i\}Ht\; over\; hbar\}right)$

is an unitary operator. It is the time evolution operator, or propagator, of a closed quantum system. If the Hamiltonian is time-independent, {U(t)} form a one parameter unitary group (more than a semigroup); this gives rise to the physical principle of detailed balance.

In many systems, two or more energy eigenstates have the same energy. A simple example of this is a free particle, whose energy eigenstates have wavefunctions that are propagating plane waves. The energy of each of these plane waves is inversely proportional to the square of its wavelength. A wave propagating in the x direction is a different state from one propagating in the y direction, but if they have the same wavelength, then their energies will be the same. When this happens, the states are said to be degenerate.

It turns out that degeneracy occurs whenever a nontrivial unitary operator U commutes with the Hamiltonian. To see this, suppose that |a> is an energy eigenket. Then U|a> is an energy eigenket with the same eigenvalue, since

- $UH\; |arangle\; =\; U\; E\_a|arangle\; =\; E\_a\; (U|arangle)\; =\; H\; ;\; (U|arangle).$

Since U is nontrivial, at least one pair of $|arang$ and $U|arang$ must represent distinct states. Therefore, H has at least one pair of degenerate energy eigenkets. In the case of the free particle, the unitary operator which produces the symmetry is the rotation operator, which rotates the wavefunctions by some angle while otherwise preserving their shape.

The existence of a symmetry operator implies the existence of a conserved observable. Let G be the Hermitian generator of U:

- $U\; =\; I\; -\; mathrm\{i\}\; epsilon\; G\; +\; O(epsilon^2)$

It is straightforward to show that if U commutes with H, then so does G:

- $[H,\; G]\; =\; 0$

Therefore,

- $$

In obtaining this result, we have used the Schrödinger equation, as well as its dual,

- $langlepsi\; (t)|H\; =\; -\; mathrm\{i\}\; hbar\; \{partialoverpartial\; t\}\; langlepsi(t)|$.

Thus, the expected value of the observable G is conserved for any state of the system. In the case of the free particle, the conserved quantity is the angular momentum.

Hamilton's equations in classical Hamiltonian mechanics have a direct analogy in quantum mechanics. Suppose we have a set of basis states $left\{left|\; n\; rightrangleright\}$, which need not necessarily be eigenstates of the energy. For simplicity, we assume that they are discrete, and that they are orthonormal, i.e.,

- $langle\; n\text{'}\; |\; n\; rangle\; =\; delta\_\{nn\text{'}\}.$

Note that these basis states are assumed to be independent of time. We will assume that the Hamiltonian is also independent of time.

The instantaneous state of the system at time t, $left|\; psileft(tright)\; rightrangle$, can be expanded in terms of these basis states:

- $|psi\; (t)rangle\; =\; sum\_\{n\}\; a\_n(t)\; |nrangle$

where

- $a\_n(t)\; =\; langle\; n\; |\; psi(t)\; rangle.$

The coefficients a_{n}(t) are complex variables. We can treat them as coordinates which specify the state of the system, like the position and momentum coordinates which specify a classical system. Like classical coordinates, they are generally not constant in time, and their time dependence gives rise to the time dependence of the system as a whole.

The expectation value of the Hamiltonian of this state, which is also the mean energy, is

- $langle\; H(t)\; rangle\; stackrel\{mathrm\{def\}\}\{=\}\; langlepsi(t)|H|psi(t)rangle$

where the last step was obtained by expanding $left|\; psileft(tright)\; rightrangle$ in terms of the basis states.

Each of the a_{n}(t)'s actually corresponds to two independent degrees of freedom, since the variable has a real part and an imaginary part. We now perform the following trick: instead of using the real and imaginary parts as the independent variables, we use a_{n}(t) and its complex conjugate a_{n}*(t). With this choice of independent variables, we can calculate the partial derivative

- $frac\{partial\; langle\; H\; rangle\}\{partial\; a\_\{n\text{'}\}^\{*\}\}$

By applying Schrödinger's equation and using the orthonormality of the basis states, this further reduces to

- $frac\{partial\; langle\; H\; rangle\}\{partial\; a\_\{n\text{'}\}^\{*\}\}$

Similarly, one can show that

- $frac\{partial\; langle\; H\; rangle\}\{partial\; a\_n\}$

If we define "conjugate momentum" variables π_{n} by

- $pi\_\{n\}(t)\; =\; mathrm\{i\}\; hbar\; a\_n^*(t)$

then the above equations become

- $$

which is precisely the form of Hamilton's equations, with the $a\_n$s as the generalized coordinates, the $pi\_n$s as the conjugate momenta, and $langle\; Hrangle$ taking the place of the classical Hamiltonian.

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Last updated on Wednesday October 01, 2008 at 08:27:56 PDT (GMT -0700)

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