动量对于时间的导数等于势能对于距离的导数

自己最近在努力学习量子力学,看到Ehrenfest’s theorem: the expected values in Quantum Mechanics obeys the classical laws. 现在证明动量对于时间的导数等于势能对于距离的导数,即:

(1)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = \langle -\dfrac{\partial V}{\partial x} \rangle \end{equation*}

First by definition, we have:

(2)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = -ih \int_{-\infty}^{\infty} \dfrac{\partial}{\partial t}(\Psi^* \dfrac{\partial \Psi}{\partial x})dx \end{equation*}

(3)   \begin{equation*} \langle \dfrac{\partial V}{\partial x} \rangle = \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} |\Psi|^2 dx \end{equation*}

How do we proof the equality? First we notice that the norm of the wave function can be expressed the following:

(4)   \begin{equation*} |\Psi|^2 = \Psi^* \cdot \Psi \end{equation*}

so (3) can be expanded as:

    \[\label{potential} \langle \dfrac{\partial V}{\partial x} \rangle = \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} \Psi^* \Psi dx \end{equation}\]

With the following integration by parts technique:

(5)   \begin{equation*} \int_{a}^{b} {u \cdot \dfrac{dv}{dx}} dx = - \int_{a}^{b} {v \cdot \dfrac{du}{dx}} dx + uv \Biggr|_{a}^{b} \end{equation*}

and also notice that at -\infty and +\infty, the wave function \Psi must be zero, so (3) can be further reduced to:

(6)   \begin{equation*} \begin{split} & \langle \dfrac{\partial V}{\partial x} \rangle \\ &= \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} \Psi^* \Psi dx \\ &= - \int_{-\infty}^{\infty} V \dfrac{\partial (\Psi^* \Psi) }{\partial x} dx + V \Psi \Psi^* \Biggr|_{- \infty}^{\infty} \\ &= - \int_{-\infty}^{\infty} V (\Psi \dfrac{\partial \Psi^*}{\partial x} + \Psi^* \dfrac{\partial \Psi}{\partial x})dx \end{split} \end{equation*}

Where should we go from here for the proof? Notice that in (2) we still have the derivative to time, so based on the basic wave function:

    \[i h \dfrac {\partial \Psi}{\partial t} = - \dfrac{h^2}{2m} \dfrac{{\partial}^2 \Psi}{{\partial} x^2}+ V \Psi\]

we can covert the derivative of \Psi and \Psi^* to time to their derivative to space as the following:

(7)   \begin{equation*} \dfrac{\partial \Psi}{\partial t} = \dfrac{i h}{2m} \dfrac{\partial^2 \Psi}{\partial x^2} - \dfrac{i}{h} V \Psi \end{equation*}

(8)   \begin{equation*} \dfrac{\partial \Psi^*}{\partial t} = - \dfrac{i h}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} + \dfrac{i}{h} V \Psi^* \end{equation*}

With these intermediate results, we can further expand the 2 as the following:

(9)   \begin{equation*} \begin{split} & \dfrac{d \langle p \rangle }{dt} \\ =& -ih \int_{-\infty}^{\infty} \dfrac{\partial}{\partial t}(\Psi^* \dfrac{\partial \Psi}{\partial x})dx \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial t} (\dfrac{\partial \Psi}{\partial x}) ] dx \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial x} (\dfrac{\partial \Psi}{\partial t}) ] dx \end{split} \end{equation*}

Applying integration by parts for the second term inside the integration above, and notice that the constant term is 0, we can have:

    \[\int_{-\infty}^{\infty} \Psi^* \dfrac{\partial}{\partial x} (\dfrac{\partial \Psi}{\partial t}) dx = - \int_{-\infty}^{\infty} \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t} dx\]

Replace this result to the above equation, we will continue to have:

    \[\dfrac{d \langle p \rangle }{dt} = -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t}] dx\]

Now, replace the time derivative of wave function with space derivative, and rearrange terms, we can have:

(10)   \begin{equation*} \begin{split} & \dfrac{d \langle p \rangle }{dt} \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t}] dx \\ &= \int_{-\infty}^{\infty} V \Psi^* \dfrac{\Partial \Psi}{\partial x} dx - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\Partial \Psi}{\partial x} \\ &+  \int_{-\infty}^{\infty} V \Psi \dfrac{\Partial \Psi^*}{\partial x} dx - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\Partial \Psi^*}{\partial x} \\ &= \langle -\dfrac{\partial V}{\partial x} \rangle - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\partial \Psi^*}{\partial x} \end{split} \end{equation*}

The last thing we need to notice is that by applying integration by parts and ignore zero constant terms, we can have:

    \[\int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} = - \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\partial \Psi^*}{\partial x}\]

so the last two terms in above equation cancel each other and we prove the desired theorem:

(11)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = \langle -\dfrac{\partial V}{\partial x} \rangle \end{equation*}