I want to ask if it is reasonable that I use the Dirac-Delta function as an initial state ($\Psi (x,0) $) for the free particle wavefunction and interpret it such that I say that the particle is exactly at x=0 during time t=0? If I use this initial state, can I also use it to predict how the wavefunction should evolve in time? That is, if $\Psi (x,0) = \delta(x) $, then, $$ \phi(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \delta(x)e^{-ikx} dx=\frac{1}{\sqrt{2 \pi}} $$ then, $$ \Psi(x,t)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \phi(k)e^{i(kx-\omega t)} dk $$ or, $$ \Psi(x,t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty}e^{i(kx-\omega t)} dk. $$

Is the above wavefunction a valid expression to see how a particle that is initially localized in the origin evolves in time?

  • $\begingroup$ possible duplicate of How do I determine the location of a free particle with Schrödinger's equation? $\endgroup$
    – alemi
    Aug 8, 2014 at 6:00
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    $\begingroup$ Note particularly the comment in the linked question: the initial conditions isn't helpful because if we have an exact position we have infinite uncertainty in momentum, and if the momentum is infinitely uncertain we can't calculate the future position. $\endgroup$ Aug 8, 2014 at 6:05
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    $\begingroup$ Hm, I only vaguely understand that qualitative argument using uncertainty principle. I guess what I want to know is a more quantitative explanation as to why this can or cannot work. $\endgroup$
    – Kurome
    Aug 8, 2014 at 6:14

3 Answers 3


That is indeed how you would go about it. Note, however, that there is nothing to guarantee that the solution is going to be reasonable, or that the integral even exists. In fact, because the Schrödinger equation is time reversible to a large extent, you are essentially guaranteed to not end up in physical states.

One thing to note is that the frequency $\omega=\omega(k)$ is a function of the wavevector $k$ through the dispersion relation, which essentially encodes the Schrödinger equation, as $\omega=E/\hbar=\hbar k^2/2m$. This means the state is \begin{align} \Psi(x,t) & = \frac{1}{2 \pi} \int_{-\infty}^{\infty}e^{i(kx-\frac{\hbar k^2}{2m} t)} dk \\ & = \frac{1}{2 \pi} e^{i\frac{m}{2\hbar t}x^2} \int_{-\infty}^{\infty} e^{-i\frac{\hbar t}{2m}(k-\frac{m}{\hbar t}x)^2} dk . \end{align} This integral, as it happens, does converge. As long as $t\neq0$, it is a Fresnel integral, and it does not need regularization to converge. (On the other hand, its convergence properties are distinct from the regularized case: it is not absolutely convergent, and the uniformity of convergence w.r.t. $x$ and $t$ is different.) Once you integrate it out, you get $$ \Psi(x,t)=\sqrt{\frac{m}{2\pi\hbar |t|}}e^{-i\mathrm{sgn}(t)\pi/4}\exp\left[i\frac{mx^2}{2\hbar t}\right]. $$ Note, in particular, that this is what you get if you plug in $a=0$ into Ruslan's initial wavefunction. That is exactly the regularization procedure which can indeed be useful but is not strictly necessary.

This state is, of course, not physical, as $|\Psi(x,t)|^2\equiv\text{const}$, but that's to be expected. What's surprising is that the amplitude is nonzero and constant for all space no matter how small $t$ is, but again that's to be expected, since $\delta(x)$ contains component at every momentum, no matter how high. This function looks as follows:

Mathematica graphics

Note that the higher-frequency components are increasingly further away from the origin. This is reasonable as these higher momenta travel faster.

Now, the real question is whether this function is actually a solution to the Schrödinger equation. It was obtained by the standard procedure in the hope that it would work, and indeed if any solution does work we expect it to be this. However, that leaves open the question of whether $$ \Psi(x,t)=\begin{cases}\delta(x) & t=0\\ \sqrt{\frac{m}{2\pi\hbar |t|}}e^{-i\mathrm{sgn}(t)\pi/4}\exp\left[i\frac{mx^2}{2\hbar t}\right]&t\neq 0\end{cases} $$ actually satisfies the differential equation $$ i\hbar\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t) $$ in any useful (presumably distributional) sense. That is left as an exercise for the reader. (Actual exercise for the reader.)

  • $\begingroup$ This is highly detailed, and even humorous, thank you. I'll try to absorb this for now and answer the 'exercise'. How do I even work out the delta function when I try to substitute it in the TISE? $\endgroup$
    – Kurome
    Aug 8, 2014 at 16:20
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    $\begingroup$ Well, that's the tough bit. If this works at all it will be in the distributional sense. You can indeed do things like differentiate delta functions - but you need to be careful about it. $\endgroup$ Aug 8, 2014 at 16:27
  • $\begingroup$ I have studied bits of the theory of distribution, do you mean I have to put in test functions in the TDSE (I mean TDSE in my previous comment)? $\endgroup$
    – Kurome
    Aug 8, 2014 at 16:51
  • $\begingroup$ Not necessarily. But you need to think very carefully about what each derivative means. $\endgroup$ Aug 8, 2014 at 16:54
  • $\begingroup$ I'm not sure how I could define a derivative on a delta distribution without test functions. It doesn't help that there's a time derivative and a space derivative on both sides of TISE. Well, I guess this is something I can think about later. $\endgroup$
    – Kurome
    Aug 8, 2014 at 17:24

Consider evolution of gaussian wave packet. Its wave function in position representation looks like:

$$\Psi(\vec r,t)=\left(\frac a{a+i\hbar t/m}\right)^{3/2}\exp\left(-\frac{\vec r\cdot \vec r}{2(a+i\hbar t/m)}\right).\tag1$$

Corresponding relative probability density is $$P(r)=|\Psi|^2=\left(\frac a{\sqrt{a^2+(\hbar t/m)^2}}\right)^3\exp\left(-\frac{a\vec r\cdot\vec r}{a^2+(\hbar t/m)^2}\right),\tag2$$

or, neglecting overall time-dependent and position-independent coefficient,

$$P'(r)=\exp\left(-\frac{a\vec r\cdot\vec r}{a^2+(\hbar t/m)^2}\right).\tag3$$

You get Dirac delta-like wave function from a initial gaussian when you take the limit $a\to0$. But for any finite $t$ the limit of $(3)$ is


i.e. at any finite time since start of evolution your position will be completely undetermined. So now nothing is really determined any more — be it momentum or position, so trying to find evolution of such a state is largely useless: you can't predict anything from your final state.


I want to ask if it is reasonable that I use the Dirac-Delta function as an intial state ($\psi(x,0)$) for the free particle wavefunction and interpret it such that I say that the particle is exactly at $x=0$ during time $t=0$?

No, because the delta function is not compliant with the Born interpretation of the function $\psi$. Evolving function that is delta function in $x$ at time $t_0$ will not give you regular wave function, but it will give you the propagator of the time-dependent Schroedinger equation. This can be used to express regular wave function at time $t$ as an integral of the wave function at some previous time $t_0$. See the section "The Free Particle Propagator" at http://physwiki.ucdavis.edu/Quantum_Mechanics/1-D_Quantum_Mechanics/Time-Dependent_Solutions%3a_Propagators_and_Representations

  • $\begingroup$ So, in other words, the state that results from propagating a delta-function initial condition is as physically reasonable as the initial condition itself. $\endgroup$ Aug 8, 2014 at 19:15
  • $\begingroup$ I would not say it that way. Delta distribution is not to be used to describe the system in the role of the $\psi$ function in the sense of the Born interpretation. It does not describe "localized system". It is just an auxiliary concept, useful for finding solution to the time-dependent Schroedinger equation. $\endgroup$ Aug 8, 2014 at 20:37
  • $\begingroup$ You mean the 'physicist's definition' for the $\delta$ function where it's infinite at x=0 and zero everywhere doesn't really work as an initial state where we can think of the particle as 'localized' at x=0? $\endgroup$
    – Kurome
    Aug 8, 2014 at 22:16
  • $\begingroup$ Oh, I wasn't implying that it's a physical state (though I disagree with such a narrow interpretation of the delta function). I simply meant that both the delta function and the evolved, Fresnel-like state have about the same level of reasonableness. I would be interested in your reasoning if you think otherwise. $\endgroup$ Aug 8, 2014 at 23:19
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    $\begingroup$ @lalala that is the first thought that comes to mind when this problem is encountered, however, it does not work mathematically, because there is no consistent theory of such thing as square root of delta distribution. My conclusion from this is that standard theory can't express such extreme state in terms of psi function that obeys the Born interpretation. Similarly for a state that has single momentum, this is also not normalizable. $\endgroup$ Jun 23, 2021 at 22:29

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