哈密顿算符

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量子哈密顿算符

量子力学的数学基础条目中介绍过，某体系的物理状态可以由一个抽象希尔伯特空间中的射线来表示。或者，在研究对象是系综时，物理状态可用一个可计数的以概率为权重的向量序列来表示。物理上的可观测量则由作用于该希尔伯特空间的自伴算符来描述。例如，一个自旋自由度为1/2的粒子对应的希尔伯特空间为 C²，一个沿某直线自由运动的粒子对应的希尔伯特空间则为Lp空间，L²(R)，the space of complex-valued and square-integrable (with respect to the Lebesgue measure) functions on the real line.

量子哈密顿算符 H 是对应于体系总能的可观测量。用数学语言来说，它是一个稠定自伴算符。如果态空间是有限空间，那么该哈密顿算符自然也是有界算符。如果态空间是无限空间，该哈密顿算符则通常无界，所以并未定义于整个空间。

物理学入门资料中常可找到下面这条公理：

哈密顿算符 H 的本征右矢(本征向量)，写作 $\left| a \right\rang$ (使用狄拉克左矢-右矢符号)，提供了一组希尔伯特空间的正交归一化的基。体系允许的能级值谱由解下列方程得到的一集本征值给出，记作 {E_a}，
$H \left| a \right\rangle = E_a \left| a \right\rangle$ 。
由于哈密顿算符 H 是厄米算符，其本征值，即能量值总是实数。

取决于体系的希尔伯特空间的性质，能量值谱可以是离散的，也可以是连续的。事实上，还有的体系在某个能量范围内值谱是连续的，而在另一个范围内则是离散的。有限势井就是一个这样的例子，兼有离散的能量为负值的态和连续的能量为正值的态。

哈密顿算符产生量子态的时间演化。如果体系在时间t时的态是 $\left| \psi (t) \right\rangle$ ，那么

$H \left| \psi (t) \right\rangle = \mathrm{i} \hbar {\partial\over\partial t} \left| \psi (t) \right\rangle$ ，

式中 $\hbar$ 是h-bar。这个方程称为薛定谔方程（它和哈密顿-雅克比方程有着相同的形式,这也是H也被称为哈密顿算符的原因之一）。给定处于初始时间(t = 0)的初始态，我们可以积分上式得到处于任何时刻的态。特别地，当 H 与时间无关时, 那么

$\left| \psi (t) \right\rangle = \exp\left(-{\mathrm{i}Ht \over \hbar}\right) \left| \psi (0) \right\rangle$ ，

式中右边的指数算符以通常的级数定义。可以证明它是一个幺正算符，也是时间演化算法的一种常见形式（亦称为传播子）。

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Energy eigenket degeneracy, symmetry, and conservation laws

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

$UH |a\rangle = U E_a|a\rangle = E_a (U|a\rangle) = H \; (U|a\rangle).$

Since U is nontrivial, at least one pair of $|a\rang$ and $U|a\rang$ 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 - iε G + O (ε 2)

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

[H, G] = 0

Therefore,

$\frac{\part}{\part t} \langle\psi(t)|G|\psi(t)\rangle = \frac{1}{\mathrm{i}\hbar} \langle\psi(t)|[G,H]|\psi(t)\rangle = 0$

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

$\langle\psi (t)|H = - \mathrm{i} \hbar {\partial\over\partial t} \langle\psi(t)|$ .

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.

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哈密顿等式

经典的哈密顿等式哈密顿力学和量子力学的哈密顿等式有着直接的类似.假设有一系列并不必要为本证态的基态 $\left\{\left| n \right\rangle\right\}$ ，由这样的类似性，我们假设这些值是离散并且是正规的， i.e.,

$\langle n' | n \rangle = \delta_{nn'}$

注意到这些基态都假设与时间无关，因而我们可以也假设哈密顿算子也独立于时间。

在时间t时，系统的暂时态 $\left| \psi\left(t\right) \right\rangle$ , 可以用以上的基态展开：

$|\psi (t)\rangle = \sum_{n} a_n(t) |n\rangle$

这里

$a_n(t) = \langle n | \psi(t) \rangle$

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

$\langle H(t) \rangle \equiv \langle\psi(t)|H|\psi(t)\rangle = \sum_{nn'} a_{n'}^* a_n \langle n'|H|n \rangle$

where the last step was obtained by expanding $\left| \psi\left(t\right) \right\rangle$ 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

$\frac{\partial \langle H \rangle}{\partial a_{n'}^{*}} = \sum_{n} a_n \langle n'|H|n \rangle = \langle n'|H|\psi\rangle$

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

$\frac{\partial \langle H \rangle}{\partial a_{n'}^{*}} = \mathrm{i} \hbar \frac{\partial a_{n'}}{\partial t}$

Similarly, one can show that

$\frac{\partial \langle H \rangle}{\partial a_n} = - \mathrm{i} \hbar \frac{\partial a_{n}^{*}}{\partial t}$

If we define "conjugate momentum" variables π_n by

$\pi_{n}(t) = \mathrm{i} \hbar a_n^*(t)$

then the above equations become

$\frac{\partial \langle H \rangle}{\partial \pi_{n}} = \frac{\partial a_{n}}{\partial t} \quad,\quad \frac{\partial \langle H \rangle}{\partial a_n} = - \frac{\partial \pi_{n}}{\partial t}$

which is precisely the form of Hamilton's equations, with the $a n$ s as the generalized coordinates, the $π n$ s as the conjugate momenta, and $\langle H\rangle$ taking the place of the classical Hamiltonian.wp:Hamiltonian_(quantum_mechanics)