📂Quantum Mechanics

The Importance of the Relative Phase of the Wave Function

설명

The wave function is often expressed as a complex exponential function as follows.

$$\psi = R e^{\i\theta}$$

At this time, in the equation, what has physical significance is not $\psi$, but $\left| \psi \right|^{2} = R^{2}$, so the value of the phase $\theta$ is not important and can be treated interchangeably.

However, the story is different when the wave function is represented as the sum of two other wave functions. In this case, the phases of each function should not be changed arbitrarily. Suppose the wave function $\psi$ is expressed as the sum of two wave functions, $\psi_{1}$ and $\psi_{2}$, as follows.

$$\psi_{1} = R_{1}e^{\i\theta_{1}}, \qquad \psi_2=R_{2}e^{\i\theta_2} \\ \psi = \psi_{1} + \psi_2$$

$$\begin{align} \left| \psi \right|^{2} = \psi^{\ast}\psi &= (\psi_{1}^{\ast}+\psi_2^{\ast})(\psi_{1}+\psi_2) \nonumber \\ &= | \psi_{1}|^{2} +|\psi_2|^{2}+ \psi_{1}^{\ast}\psi_2+\psi_2^{\ast}\psi_{1} \nonumber \\ &= {R_{1}}^{2}+{R_{2}}^{2}+R_{1}R_{2}e^{\i(\theta_2-\theta_{1})}+R_{1}R_{2}e^{\i(\theta_{1}-\theta_2)} \nonumber \\ &= {R_{1}}^{2}+{R_{2}}^{2}+R_{1}R_{2} \color{blue}{\left[ e^{\i(\theta_2-\theta_{1})}+e^{\i(\theta_{1}-\theta_2)} \right]} \nonumber \\ &= {R_{1}}^{2} + {R_{2}}^{2} +2R_{1}R_{2}\cos(\theta_{1}-\theta_2) \end{align}$$

When looking at $(1)$, because it contains $\theta_{1} - \theta_{2}$ in the equation, you can see that the phases of each wave function cannot be changed. The solution for the blue part is as follows. According to the Euler’s formula, $e^{i\theta}=\cos\theta+i\sin\theta$,

$$\begin{align*} &\quad\ e^{\i(\theta_2-\theta_{1})}+e^{\i(\theta_{1}-\theta_2)} \\ &= \cos (\theta_2-\theta_{1})+\i\sin (\theta_2-\theta_{1})+\cos (\theta_{1}-\theta_2)+\i\sin (\theta_{1}-\theta_2) \\ &= [ \cos (\theta_2-\theta_{1})+\cos (\theta_{1}-\theta_2)]+[\i\sin (\theta_2-\theta_{1})+\i\sin (\theta_{1}-\theta_2)] \\ &= 2\cos(\theta_{1}-\theta_2) \end{align*}$$