Exploring Quantum Physics - Week 7 Question 10

Question:

Suppose we have a wave function of an electron under an electric field, what would the wave function be in the anti technique we just derived?

Solution:

We just simply do $\alpha\beta$ to both side of the equation, that gives.

$\begin{eqnarray*} i\hbar \frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - qF\hat{x} + mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ -i\hbar \frac{\partial \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - QF\hat{x} + Mc^2 \hat{\beta}\right) \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - QF\hat{x} + Mc^2 \hat{\beta}\right) \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\left( c \hat{\alpha} \hat{p} - QF\hat{x} - Mc^2 \hat{\beta}\right) \hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\hat{\beta} \left( -c \hat{\alpha} \hat{p} - QF\hat{x} - Mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\hat{\beta} \left( c \hat{\alpha} \hat{p} + QF\hat{x} + Mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ \end{eqnarray*}$

So the new 'particle' has the charge flipped and the mass stay the same - that envisions positron!