In the above figure, the beamsplitter puts the electron in a superposition of taking the upper path and taking the lower path. In both paths, when the electron gets to the mirror, it is stopped and held there. During that time when the electron is held in place at a mirror, 2 electric charges each with charge Q are brought near the upper mirror in a symmetric manner such that the net electric field caused by the 2 charges at the upper mirror is 0. We assume that the lower mirror is far enough away from the upper mirror such that the electric potential (and electric field) caused by the 2 charges is 0 at the lower mirror. So, this creates an electric potential difference between upper and lower mirrors equal to Δ Δ U = 2 Q 4 π π ϵ ϵ 0 r , where r is the distance of the charges from the mirror and ϵ ϵ 0 is the electric constant. The electron is held there for a time T , after which the charges are moved away and the electron is allowed to continue moving along its path. Assuming that the time we take to move the 2 charges to and from the mirror is much smaller than T , this time that the electron spends at the mirror causes a phase shift equal to