# Stationary states in Schrödinger's equation

I was following my book's analysis on a free particle possessing potential energy and it got to the point where $$(\psi(x,t))^2$$ no longer depended on time. Maybe I'm misunderstanding but does this imply that the particle is not moving? If this is the case, how? Doesn't Schrödinger's equation also include kinetic energy?

The long answer: Suppose you have a ring rotating, so that if you look from far away the ring appears ''stationary''. However, each part of the ring is still moving, despite the fact that the mass distribution is constant.

It is the same with stationary states. Take the free particle as an example. The stationary states are of the form $\psi\sim e^{i (-\omega t+kx)}$. So, graphing this wave function, you will see that it is indeed moving.

The measure for the moving of particles is the "current" $j=\frac{\hbar}{2mi}(\psi^*\partial_x\psi-\psi\partial_x\psi^*)$. You can try calculating this on the free particle, and it should be a nonzero number.

What book are you reading? It's helpful to know so that I know what formalism you're familiar with. Examing Schrodinger's equation,

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

where $$H$$ is the Hamiltonian, $$K$$ the kinetic energy, and $$V$$ the potential energy, we see that in the "free particle" case, when $$V = 0$$, the energy is determined only by the kinetic energy. In nature, potentials can be very complicated, so textbooks of course begin with simple examples, i.e. infinite square well, finite steps, harmonic oscillator, etc...

To find the state of the system, represented by the wavefunction $$\Psi(x,t)$$, we need to solve Schrodinger's equation. Some potentials are simple enough that they are "separable" meaning they can be separated into a product of functions of their respective variables. In intro quantum mechanics, the Schrodinger equation is a separable differential equation $$\underline{when}$$ the potential is independent of time. All (or most) of the examples in your textbook have potentials that are independent of time, and they are all probably solved by the method of separation of variables, thus they all have solutions called "stationary states."

They are called stationary because they do not depend on time, rather they depend on position, and they are eigenstates of the Hamiltonian (they satisfy the eigenvalue equation for $$H$$). What does this mean? Stationary states are nice because they 1) provide time independent probability densities and expectation values, 2) they are states of definite total energy, 3) the general solution of this separable Schrodinger equation is a linear combination of the stationary states.

So then, since stationary states form a complete orthonormal basis, we may write,

$$\Psi(x) = \sum_{n = 0}^{\inf} c_{n} \psi(x).$$

where the $$c_{n}$$ are constants determined from an orthonormality condition. To obtain the full time dependent solution, we simply multiply the stationary states above by the time dependent factor obtained by solving the time dependent part of the Schrodinger equation:

$$\Psi(x,t) = \sum_{n = 0}^{\inf} c_{n} \psi(x) e^{\frac{-iE_{n}}{\hbar}t}.$$

The classical notion of the particle "moving" is nonsense in quantum mechanics. There is a probability density that it will be found somewhere, and it will have an expectation value (the average over a large amount of identically prepared systems) of position from which you can obtain an expectation value of velocity, but if it pleases you, one may of course consider the particle as moving between states that it's measured to be in at various times, or the more traditional attitude is that it's nowhere - it's just a probability density - until it's measured. Examine a stationary state as it evolves in time here: since $$\rho$$ is constant in time, the area under the curve is constant.

For the correspondence between the classical notions of velocity and momentum to quantum mechanics, see Ehrenfest's Theorem.

• Awesome! So I assume in some instances that the expected value of x can change over time. If so, if you were to take the derivative of the expected value with respect to time, would this be the expected value of the velocity? I am reading "Modern Physics" by Randy Harris, pretty intro stuff. Commented Aug 17, 2018 at 2:11
• If you are taking an expectation value of some observable using a stationary state, then the expectation value is constant in time. Period. However, if you are taking an expectation value of some observable using a linear combination of stationary states, then the expectation value will depend on time (due to cross term). So, the general solution I wrote above (with the sum) does produce time dependent expectation values. Griffith's intro to quantum mechanics book is a great place to start, and in it he talks about Ehrenfest theorem, for which i added a link in my answer at the bottom. Commented Aug 17, 2018 at 2:38