Chapter 1 - Section 1.5 - Momentum - Problems - Page 18: 1.8

$\text{1) Adding a constant }V_0\text{ to potential energy term just adds }$ $\text{a phase factor of }e^{-\iota V_0 t/\hbar}$ $$\psi_0 = \psi^{-\iota V_0 t/\hbar}$$ $\text{2) Adding a constant }V_0\text{ to potential energy term doesn't }$ $\text{have any effect on expectation value of a dynamic variable.}$

$\text{Suppose that } \psi\text{ is the solution to the time-dependent}$ $\text{Schrödinger equation without the added potential, So}$ $$\iota \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2}+V\psi\qquad ...(1)$$ $\text{Claim that the solution to the Schrödinger Equation with the }$ $\text{constant potential offset } V_0 \text{ should be:}$ $$\psi_0 = \psi e^{-\iota V_0 t/\hbar}$$ $\text{Consider,}$ $$\iota \hbar \frac{\partial \psi_0}{\partial t} = \iota \hbar \frac{\partial \psi}{\partial t} e^{-\iota V_0 t/\hbar}+\iota \hbar\big(\frac{-\iota V_0}{\hbar}\big) \psi e^{-\iota V_0t/h}$$ $\text{From equation (1)}$ $$\iota \hbar \frac{\partial \psi_0}{\partial t} = \big(-\frac{\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2}+V\psi\big) e^{-\iota V_0 t/\hbar}+V_0 \psi e^{-\iota V_0t/h}$$ $\text{Also,}$ $$\frac{\partial^2 \psi_0}{\partial x^2} = \frac{\partial^2 \psi}{\partial x^2}e^{-\iota V_0 t/\hbar}$$ $\text{Therefore, equation becomes:}$ $$\iota \hbar \frac{\partial \psi_0}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 \psi_0}{\partial x^2}+V\psi e^{-\iota V_0 t/\hbar}+V_0 \psi e^{-\iota V_0t/h}$$ $$\iota \hbar \frac{\partial \psi_0}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 \psi_0}{\partial x^2}+(V+V_0) \psi e^{-\iota V_0t/h}$$ $$\iota \hbar \frac{\partial \psi_0}{\partial t} = -\frac{\hbar}{2m}\frac{\partial^2 \psi_0}{\partial x^2}+(V+V_0) \psi_0$$ $\psi_0 \text{ satisfies the Schrödinger wave equation.}$ $\text{Thus, shifting the overall potential energy by a constant }$ $\text{amount }V_0\text{ just adds a phase factor }e^{-\iota V_0t/\hbar}$ $\text{Expectation Value of any dynamic value }Q(x, p)$ $\text{is given by }$ $$\langle Q(x, p) \rangle = \int \psi^* Q\big(x, \frac{\hbar}{\iota}\frac{\partial}{\partial x}\big)\psi dx$$ $\text{For }\psi = \psi_0$ $$\langle Q(x, p) \rangle = \int \psi_0^* Q\big(x, \frac{\hbar}{\iota}\frac{\partial}{\partial x}\big)\psi_0 dx$$ $$\langle Q(x, p) \rangle = \int \psi^*e^{\iota V_0 t/\hbar} Q\big(x, \frac{\hbar}{\iota}\frac{\partial}{\partial x}\big)\psi e^{-\iota V_0 t/\hbar} dx$$ $$\langle Q(x, p) \rangle = \int \psi^* Q\big(x, \frac{\hbar}{\iota}\frac{\partial}{\partial x}\big)\psi dx$$ $\text{So adding a constant }V_0\text{ to potential energy term doesn't }$ $\text{have any effect on expectation value of a dynamic variable.}$