Why is a wave function time dependent? And what does it physically mean when the probability density density function is varying with time. I'm not getting how to physically interpret that. Will taking example of a particle in 1d box with superposition state two energy levels help? Because it's simplest system(I can think of) whose probability density is time dependent.

Edit: I'm not looking for a proof or a mathematical explanation of how it depends on time but I'm questioning why should the wavefunction depend on time. I am getting answers derived from time dependent Schrodinger's equation but then we are already assuming that wave function time dependent. So to be more clear my intent was question "where is the time dependence of Schrodinger's equation coming from? Where did the time dependence come from?"

Edit: I actually started questioning about why the probability density of superposition of states is time dependent, on second look I realised it came from the time dependence of wave function and that again was because of time dependence term in Schrodinger's equation.

Probability density<--wave function<---schrodingers equation<--- [???????]

Is it the end or any more layers are there? Please go down as many layers as possible.

  • 2
    $\begingroup$ "Why is a wave function time dependent?" - Is it just that (1) expectation values, in general, evolve with time and (2) in the Schrodinger picture, the states 'carry' the time dependence? (In the Heisenberg picture, the states are time independent and so the operators carry the time dependence). $\endgroup$ Dec 5, 2017 at 14:12
  • $\begingroup$ 1) The expectation values are calculated over a period of infinite time right? How would they evolve over time? Anyway that doesn't actually answer my question of why the wavefunction is timedependent because $\endgroup$
    – dushyanth
    Dec 5, 2017 at 14:18
  • $\begingroup$ 2) but that still means that probability density function is time dependent right? $\endgroup$
    – dushyanth
    Dec 5, 2017 at 14:20
  • 1
    $\begingroup$ Wavefunction is time dependent because of $\partial_t$ in Schroedinger equation. Probability density is time dependent because you integrate out spatial coordinates, not the time. $\endgroup$
    – Ice-Nine
    Dec 5, 2017 at 14:23
  • 2
    $\begingroup$ @dushyanth No. In quantum mechanics expectation values are taken by repeating the same experiment many times and averaging the results. They are not time averages (like you get in statistical mechanics) $\endgroup$ Dec 5, 2017 at 14:24

2 Answers 2


Any equation claiming to describe fundamental process has to have time. We know that in nature system changes with time. They are not static. For example, absorption of a photon by a hydrogen atom is a dynamic process and we need to have a theory which can predict this dynamical process in time. Since Schrodinger equation is supposed to be an equation describing nature, time needs to get involved somewhere.

The question then is why did Schrodinger choose this particular equation? There is no rigorous argument to that. If you read Schrodinger's original paper, you will see that Schrodinger used Hamilton Jacobi Equation, $H(q,\frac{\partial S}{\partial q})-\frac{\partial S}{\partial t}=0$. He replaced action $S$ by, $$\psi=e^{KS},$$ where $K$ is some constant with the dimension of action (seems familiar?). He then postulates that integral of the left-hand side of above equation should be extremized. This leads to Schrodinger Equation. The factor of time derivative in Schrodinger equation comes from the term $\frac{\partial S}{\partial t}$ in Hamilton Jacobi Equation.

Above procedure may seem, and is, ad-hoc. But he was able to calculate energy spectrum of Hydrogen atom successfully with it. He also wrote a paper in which he sort of gives motivation for this procedure. He used the geometrical formulation of Hamilton Jacobi Equation and argued that quantum physics differ from classical in the same way wave optics differ from geometric optics.

In short, we require wave function to have time dependence if it has to describe nature. The way this dependence was introduced by Schrodinger is little bit shaky but not entirely bogus. I would suggest you to read his original papers if you want to understand his motivation.


Wavefunctions can be time-dependent because the time-dependent Schrodinger equation is linear: the sum of two time-dependent solutions is also a solution. Thus the general solution will be a sum of time-dependent solutions $$ \Psi(x,t)=\sum_k a_k\psi_k(x)e^{-i E_k t/\hbar} \tag{1} $$ where the coefficients $a_k$ are determined from the initial conditions.

In many practical situations, the states of interest are not eigenstates of the Hamiltonian because of initial conditions (or alternatively because of measurements). A famous example is the tunnelling of nitrogen atoms through the hydrogen plane in an ammonia molecule (see this discussion of ammonia inversion). In this case the nitrogen atom is found to oscillate "above" or "below" the hydrogen plane: the probability density is certainly time-dependent.

For a particle in a box, an example might be a system described at $t=0$ by a wave function initially heavily concentrated in the left half of the well of width $L=1$: $$ \Psi(x,0)= \sqrt{858}x(x-1)^5\, ,\tag{2} $$ with initial probability distribution given by the figure below: enter image description here

Expanding (1) in eigenstates of the square well will produce a sum of the type (1). The first few $a_k$'s are $$ \begin{array}{l} a_1=-\frac{20 \sqrt{429} \left(\pi ^2-12\right)^2}{\pi ^7}=-0.622487\, , \\ a_2=-\frac{5 \sqrt{429} \left(\pi ^2-6\right)}{2 \pi ^5}=-0.654766\, , \\ a_3=-\frac{20 \sqrt{\frac{143}{3}} \left(4-3 \pi ^2\right)^2}{81 \pi ^7}=-0.370154\, ,\\ a_4=-\frac{5 \sqrt{429} \left(2 \pi ^2-3\right)}{32 \pi^5}=-0.177025\, ,\\ a_5=-\frac{4 \sqrt{429} \left(12-25 \pi ^2\right)^2}{15625 \pi ^7}=-0.0967373\, . \end{array} $$ The sum of (1) extends to $\infty$ but already the sum $\sum_{k=1}^4 a_k^2=0.985$ indicating that all other eigenstates have a small contribution to the full wavefunction and thus do not affect much the evolution of the full wavefunction.

  • $\begingroup$ Please look at the edit in question details.. sorry if my framing of question was not clear before. $\endgroup$
    – dushyanth
    Dec 5, 2017 at 17:41
  • $\begingroup$ I have just made 2nd new edit. Hope I'm clear this time $\endgroup$
    – dushyanth
    Dec 5, 2017 at 17:51
  • $\begingroup$ @dushyanth I'm not sure I understand you question then. By simple separation of variable one initially searches for $\Psi_n(x,t)=\psi_n(x)e^{-iE_nt/\hbar}$ so the time dependence enters at that stage. It's only when the wavefunction at $t=0$ is one of the $\Psi_n(x,t)$ that the time dependence in the probability density will go away. In general however, your initial wavefunction will NOT be of this sort, as illustrated with an example in my answer. It's easy to see that, in the example above, $\vert \Psi(x,t)\vert^2$ is time dependent since all $\psi_n(x)$ have a different $t4-dependence. $\endgroup$ Dec 5, 2017 at 17:52
  • $\begingroup$ Please take a look at the second edit $\endgroup$
    – dushyanth
    Dec 5, 2017 at 17:56
  • 1
    $\begingroup$ @dushyanth No problem... hopefully it will become clearer in your mind and you can in time refine your questions. $\endgroup$ Dec 5, 2017 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.