I am reading Griffith's Intro to Quantum Mechanics and when explaining the Time Independent Schrodinger Equation, he says the general solution to $d\varphi/dt$ is:

$$\varphi (t)=e^{-iEt/h}$$

Even though he says it's a result of just multiplying through by dt and integrating, I don't see where the exponential is coming from. I know waves can be represented using complex exponential but I don't see how it is just plugged in as a general solution to finding $\varphi$ with $E$ involved.


2 Answers 2


It is a simple first order ODE, the solution is straightforward. From separating the variables of the original Schrödinger PDE you get for the time-dependent part:

$$\frac {1}{\phi(t)} \frac {d\phi(t)}{dt} =\frac{-iE}{\hbar} \tag 1$$ Noticing that the left side can be simplified by chain rule: $$\frac {1}{\phi(t)} \frac {d\phi(t)}{dt} =\frac d{dt}ln(\phi(t)) $$

Then, integrating both sides of equation $(1)$:

$$\int \frac d{dt}ln(\phi(t))dt=\int \frac{-iE}{\hbar}dt \implies \phi(t)=Ce^{-iEt/\hbar}$$


I don't see where the exponential is coming from

First, recall that the solutions to the Time Independent Schrodinger Equation (TISE) are the spatial part of the solutions to the Time Dependent Schrodinger Equation (TDSE) which are separable (the product of a function of the spatial coordinates only and a function of time only):

$$\Psi(x,t) = \psi(x)\phi(t)$$

where $\psi(x)$ is an eigenfunction of $\hat{H}$

$$\hat{H}\psi(x) = E\psi(x)$$

Then the TDSE

$$\hat{H}\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$


$$\hat{H}\psi(x)\phi(t) = E\psi(x)\phi(t) = i\hbar\frac{\partial}{\partial t}\psi(x)\phi(t)$$

and so

$$i\hbar\frac{d\phi(t)}{dt} = E\phi(t)$$

It's easy to check that $\phi(t) = e^{-iEt/\hbar}$ solves this equation

$$i\hbar\frac{d}{dt}e^{-iEt/\hbar} = i\hbar \left(-iE/\hbar\right)e^{-iEt/\hbar} = Ee^{-iEt/\hbar} = E \phi(t)$$


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