# Determination of a wave function

How do you determine the quantum wave function for a specific case ? I had a course about quantum mechanics, but I believe we never discussed how you get a specified wave function for a specified case. I am really curious about how this is done. I know how to work with wave functions, but where do they come from...

• I'm not sure ai understand your question, Presumably you would solve the time dependent Schrodinger equation for this specific potential, and its solutions would be the spatial wave functions $\psi(x)$ – ZeroTheHero Mar 2 '17 at 0:33
• Can I suggest a book that explains what you want to know by worked example: Quantum Mechanics Demystified by David McMahon. Look it up on amazon and read the TOC. – user146020 Mar 2 '17 at 0:36
• Course in QM usually include solving the Schrodinger Equation for simple scenarios like a Particle in a Box. Such examples are worked through in elementary QM textbooks - and even in general undergraduate texts. – sammy gerbil Mar 2 '17 at 1:56

In general, the way to find a wavefunction is using the time-dependent Schrodinger equation : $$i\hbar\frac{d\left|\psi\right>}{dt} = H\left|\psi\right>$$
Where H is the hamiltonian operator : it acts on the wavefunction $\left|\psi\right>$. Now, H is what will be the "equivalent" of forces in the Newton equation $\vec{F} = \frac{d\vec{p}}{dt}$, and it is going to depend on the problem you are considering.
I'm assuming you are not familiar with "bra" and "kets", and that you worked only with "$\psi(x,t)$" as is often the case in an introductory course. Although the equation I gave you is more general, we can re-express it in the "x" space as such : $$i\hbar\frac{d\psi}{dt} = H\psi$$ Not so much different, but now H is a differential operator acting of $\psi(x,t)$. For a free particle, for example, we have $H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$
Finally, we often use a separation of variables $\psi(x,t) = T(t)\psi(x)$ to separate time and spatial part. This gives 2 equations :$$i\hbar T'(t) = ET(t)$$ $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{d x^2} = E\psi(x)$$