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Question

Consider a freely moving quantum particle with mass m and speed v. Its energy is E=K+U=12mv2+0E=K+U=\frac{1}{2} m v^{2}+0. Determine the phase speed of the quantum wave representing the particle and show that it is different from the speed at which the particle transports mass and energy.

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As we know, the particle has energy :

E=K+UE=12mv2+0E=K=12mv2E=hfλ=hmvvp=fλ\begin{align*} E&=K+U\\ E&=\frac{1}{2}mv^2+0\\ E&=K=\frac{1}{2}mv^2\\ E&=hf \\ \Rightarrow \lambda&=\frac{h}{mv} \tag{Wavelength.}\\ \Rightarrow v_{p}&=f \lambda \tag{Phase velocity $\rightarrow$ equation 1.}\\ \end{align*}

Now, we have to substitute values in equation 1 and solve this equation :

vp=fλvp=Ehhmvvp=12mv2hhmvvp=mv22hhmvvp=v2v \begin{align*} v_{p}&=f \lambda \tag{Equation 1.}\\ v_{p}&=\frac{E}{h} \cdot \frac{h}{mv} \\ v_{p}&=\frac{\frac{1}{2} mv^2}{h} \cdot \frac{h}{mv} \tag{Substitute values in equation.}\\ v_{p}&=\frac{mv^2}{2h} \cdot \frac{h}{mv} \\ v_{p}&=\frac{v}{2} \ne v \\\ \end{align*}

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