Expectation Values
In classical mechanics, the solution to an equation of motion is a function of a measurable quantity, such as \(x(t)\), where \(x\) is the position and \(t\) is the time. Note that the particle has one value of position for any time \(t\). In quantum mechanics, however, the solution to an equation of motion is a wavefunction, \(\Psi \, (x,t)\). The particle has many values of position for any time \(t\), and only the probability density of finding the particle, \(|\Psi \, (x,t)|^2\), can be known. The average value of position for a large number of particles with the same wavefunction is expected to be
\[\langle x \rangle = \int_{-\infty}^{\infty} xP(x,t) \, dx = \int_{-\infty}^{\infty} x \Psi^* (x,t) \, \Psi \, (x,t) \, dx. \label{7.6} \]
This is called the expectation value of the position. It is usually written
\[\langle x \rangle = \int_{-\infty}^{\infty} \Psi^* (x,t) \, x \Psi \, (x,t) \, dx. \label{7.7} \]
where the \(x\) is sandwiched between the wavefunctions. The reason for this will become apparent soon. Formally, \(x\) is called the position operator.
At this point, it is important to stress that a wavefunction can be written in terms of other quantities as well, such as velocity (\(v\)), momentum (\(p\)), and kinetic energy (\(K\)). The expectation value of momentum, for example, can be written
\[\langle p \rangle = \int_{-\infty}^{\infty} \Psi^* (p,t) \, p\Psi \, (p,t) \, dp, \label{7.8} \]
where \(dp\) is used instead of \(dx\) to indicate an infinitesimal interval in momentum. In some cases, we know the wavefunction in position, \(\Psi \, (x,t)\), but seek the expectation of momentum. The procedure for doing this is
\[\langle p \rangle = \int_{-\infty}^{\infty} \Psi^* (x,t) \, \left(-i\hbar \dfrac{d}{dx}\right) \, \Psi \, (x,t) \, dx, \label{7.9} \]
where the quantity in parentheses, sandwiched between the wavefunctions, is called the momentum operator in the x-direction. [The momentum operator in Equation \ref{7.9} is said to be the position-space representation of the momentum operator.] The momentum operator must act (operate) on the wavefunction to the right, and then the result must be multiplied by the complex conjugate of the wavefunction on the left, before integration. The momentum operator in the x-direction is sometimes denoted
\[\langle p \rangle = - i\hbar \dfrac{d}{dx},\label{7.10} \]
Momentum operators for the y- and z-directions are defined similarly. This operator and many others are derived in a more advanced course in modern physics. In some cases, this derivation is relatively simple. For example, the kinetic energy operator is just
\[\begin{align} (K)_{op} &= \dfrac{1}{2}m(v_x)_{op}^2 \\[5pt] &= \dfrac{(p_x)^2_{op}}{2m} \\[5pt] &= \dfrac {\left(-i\hbar \dfrac{d}{dx}\right)^2}{2m} \\[5pt] &= \dfrac{-\hbar^2}{2m} \left(\dfrac{d}{dx}\right)\left(\dfrac{d}{dx}\right).\label{7.11} \end{align} \]
Thus, if we seek an expectation value of kinetic energy of a particle in one dimension, two successive ordinary derivatives of the wavefunction are required before integration.
Symmetry can simplify calculations
Expectation-value calculations are often simplified by exploiting the symmetry of wavefunctions. Symmetric wavefunctions can be even or odd. An even function is a function that satisfies
\[\psi(x) = \psi(-x). \label{7.12} \]
In contrast, an odd function is a function that satisfies
\[\psi(x) = -\psi(-x).\label{7.13} \]
An example of even and odd functions is shown in Figure \(\PageIndex{7}\). An even function is symmetric about the y-axis. This function is produced by reflecting \(\psi (x)\) for \(x > 0\) about the vertical y-axis. By comparison, an odd function is generated by reflecting the function about the y-axis and then about the x-axis. (An odd function is also referred to as an anti-symmetric function.)
图\(\PageIndex{7}\)：偶数和奇数波函数的示例。
通常，偶数函数乘以偶数函数会生成偶数函数。 偶数函数的一个简单示例是乘积\(x^2e^{-x^2}\)（偶数倍即偶数）。 同样，奇数函数乘以奇数函数会生成偶数函数，例如 x sin x（奇数次为偶数）。 但是，奇数函数乘以偶数函数会生成奇数函数，例如\(x^2e^{-x^2}\)（奇数倍偶数就是奇数）。 奇数函数在所有空间上的积分为零，因为该函数在 x 轴上方的总面积会抵消其下方的（负）区域。 如下一个示例所示，奇数函数的这个属性非常有用。
示例\(\PageIndex{2A}\): Expectation Value (Part I)
粒子的归一化波函数是
\[\psi(x) = e^{-|x|/x_0} /\sqrt{x_0}. \nonumber \]
找出持仓的期望值。
策略
将波函数替换为方程\ ref {7.7} 并进行计算。 位置运算符仅引入乘法因子，因此位置运算符不必被 “夹住”。
解决方案
先乘以，然后整合：
\[\begin{align*} \langle x \rangle &= \int_{-\infty}^{\infty} dx\,x|\psi(x)|^2 \nonumber \\[4pt] &= \int_{-\infty}^{\infty} dx\, x|\dfrac{e^{-|x|/x_0}}{\sqrt{x_0}}|^2 \nonumber \\[4pt] &= \dfrac{1}{x_0} \int_{-\infty}^{\infty} dx\, xe^{-2|x|/x_0} \nonumber \\[4pt] &= 0. \nonumber \end{align*} \nonumber \]
意义
integrand (\(xe^{-2|x|/x_0}\)) 中的函数是奇数，因为它是奇数函数 (x) 和偶数函数 (\(e^{-2|x|/x_0}\)) 的乘积。 积分消失是因为函数围绕 x 轴的总面积抵消了其下方的（负）区域。 结果 (\(\langle x \rangle = 0\)) 并不奇怪，因为概率密度函数大约是对称的\(x = 0\)。
示例\(\PageIndex{2B}\): Expectation Value (Part II)
限制在 0 和 L 之间的区域的粒子随时间变化的波函数为
\[\psi(x,t) = A \, e^{-i\omega t} \sin \, (\pi x/L) \nonumber \]
其中\(\omega\)是角频率，\(E\)是粒子的能量。 （注意：由于限制（0 到 L），该函数随正弦变化。 W\(x = 0\) hen，正弦系数为零，波函数为零，这与边界条件一致。） 计算位置、动量和动能的预期值。
策略
我们必须首先对波函数进行归一化才能找到 A。 然后我们使用运算符来计算期望值。
解决方案
计算归一化常量：
\[\begin{align*} 1 &= \int_0^L dx\, \psi^* (x) \psi(x) \nonumber \\[4pt] &= \int_0^L dx \, \left(A e^{+i\omega t} \sin \, \dfrac{\pi x}{L}\right) \left(A e^{-i\omega t} \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= A^2 \int_0^L dx \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= A^2 \dfrac{L}{2} \nonumber \\[4pt] \Rightarrow A &= \sqrt{\dfrac{2}{L}}. \nonumber \end{align*} \nonumber \]
头寸的期望值为
\[\begin{align*}\langle x \rangle &= \int_0^L dx \, \psi^* (x) x \psi(x) \nonumber \\[4pt] &= \int_0^L dx \, \left(A e^{+i\omega t} \sin \, \dfrac{\pi x}{L}\right) x \left(A e^{-i\omega t} \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= A^2 \int_0^L dx\,x \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= A^2 \dfrac{L^2}{4} \nonumber \\[4pt] \Rightarrow A &= \dfrac{L}{2}. \nonumber \end{align*} \nonumber \]
x 方向动量的期望值也需要积分。 要设置这个积分，关联的运算符必须（按照规则）在波函数上向右移动\(\psi(x)\)：
\[\begin{align*} -i\hbar\dfrac{d}{dx} \psi(x) &= -i\hbar \dfrac{d}{dx} Ae^{-i\omega t}\sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= - i\dfrac{Ah}{2L} e^{-i\omega t} \cos\, \dfrac{\pi x}{L}. \nonumber \end{align*} \nonumber \]
因此，动量的预期值为
\[ \begin{align*} \langle p \rangle &= \int_0^L dx \left(Ae^{+i\omega t}sin \dfrac{\pi x}{L}\right)\left(-i \dfrac{Ah}{2L} e^{-i\omega t} cos \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= -i \dfrac{A^2h}{4L} \int_0^L dx \, \sin \, \dfrac{2\pi x}{L} \nonumber \\[4pt] &= 0. \nonumber \end{align*} \nonumber \]
积分中的函数是一个波长等于井宽的正弦函数，L —大约是奇数函数\(x = L/2\)。 结果，积分消失了。
x 方向动能的预期值要求关联的运算符对波函数起作用：
\[ \begin{align} -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2} \psi (x) &= - \dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} Ae^{-i\omega t} \, \sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= - \dfrac{\hbar^2}{2m} Ae^{-i\omega t} \dfrac{d^2}{dx^2} \, \sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= \dfrac{Ah^2}{8mL^2} e^{-i\omega t} \, \sin \, \dfrac{\pi x}{L}. \nonumber \end{align} \nonumber \]
因此，动能的期望值为
\[\begin{align*} \langle K \rangle &= \int_0^L dx \left( Ae^{+i\omega t} \, \sin \, \dfrac{\pi x}{L}\right) \left(\dfrac{Ah^2}{8mL^2} e^{-i\omega t} \, \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= \dfrac{A^2h^2}{8mL^2} \int_0^L dx \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= \dfrac{A^2h^2}{8mL^2} \dfrac{L}{2} \nonumber \\[4pt] &= \dfrac{h^2}{8mL^2}. \end{align*} \nonumber \]
意义
处于该状态的大量粒子的平均位置为\(L/2\)。 这些粒子的平均动量为零，因为给定粒子向右或向左移动的可能性相同。 但是，粒子没有处于静止状态，因为它的平均动能不为零。 最后，概率密度为
\[|\psi|^2 = (2/L) \, \sin^2 (\pi x/L). \nonumber \]
该概率密度在位置\(L/2\)上最大，在\(x = 0\)和处均为零\(x = L\)。 请注意，这些结论并不明确取决于时间。
练习\(\PageIndex{3}\)
对于上述示例中的粒子，求出将其定位在位置\(0\)和之间的概率\(L/4\)。
回答
\((1/2 - 1/\pi) /2 = 9\%\)
量子力学做出了许多令人惊讶的预测。 但是，在1920年，尼尔斯·玻尔（哥本哈根尼尔斯·玻尔研究所的创始人，我们从中得到 “哥本哈根解释” 一词）断言，量子力学和经典力学的预测必须与所有宏观系统一致，例如绕行星运行、弹跳球、摇椅和弹簧。 这种对应原则现已得到普遍接受。 它表明经典力学的规则近似于能量非常大的系统的量子力学规则。 量子力学描述了微观和宏观世界，但经典力学只描述了后者。