# Evolution of Delta Function in Free Particle Hamiltonian

I was curious what would happen if we had a free particle hamiltonian and saw what would happen if your initial wavefunction was a delta function and time evolve it using Schrodinger's Equation (say in one dimension). Here were my thoughts:

$$-\frac{\hbar^2}{2m}\partial^2_x\psi = i\hbar \partial_t \psi$$

Taking the Fourier transform, which I call $$F(\cdot)$$, and $$F(\psi) = \tilde{\psi}$$:

$$\partial_t \tilde{\psi} = \frac{\hbar i}{2m} (2\pi i k)^2 \tilde{\psi}$$

$$\implies \tilde{\psi} = \tilde{\psi}(t=0)\exp(\frac{-2\pi^2\hbar k^2 it}{m})$$

If we our initial wavefunction is a delta function (say centered at $$x=0)$$:

$$\tilde{\psi}(t=0) = F(\delta(x)) = \int \delta(x) \exp(-2\pi i xk) dx= 1$$

$$\implies \tilde{\psi}(k, t) = \exp(\frac{-2\pi^2\hbar k^2 it}{m})$$

Thus, in position space:

$$\psi(x, t) = F(\tilde{\psi}) = \int \exp(\frac{-2\pi^2\hbar k^2 it}{m}) \exp(2\pi ixk)dk$$

I believe this is correct, and (if it is) my main question is how do you interpret it?

• Did you try to evaluate the last Gaussian integral? – Qmechanic Aug 27 at 14:27
• Your maximally localized function in position space is a function that is maximally delocalized in momentum space, which is why you get a constant function of k when you transform it to k space, which is nothing else but momentum space except some constants. The time evolution shows that each momentum component of your function has a phase factor that oscillates with the corresponding energy $E(k)=(\hbar k)^2/2m$. It is important to remember that these states are idealized and cannot be realized as physical state. They are useful mathematical tools but physically impossible. – Hans Wurst Aug 27 at 14:43

You've found the integral representation of Green's function for the free particle Schrodinger equation. Delta-function is not a square-integrable function, but in quantum mechanics we often consider $$\delta(x)$$ as the eigenfunction of the operator $$\hat{x}$$ in coordinate representation: $$\hat{x}|x'\rangle = x'|x'\rangle, \quad \langle x| x'\rangle = \delta(x-x').$$ From your question formulation follows equality $$\psi(x,t) = \langle x | e^{-\frac{i}\hbar \hat{H} t} | x' = 0 \rangle \qquad (*)$$ Free particle problem possess translational invariance, so $$\psi(x,t) = \langle x+x_0 | e^{-\frac{i}\hbar \hat{H} t} | x_0 \rangle$$ for any $$x_0$$. As $$\psi(x,t)$$ is the matrix element of the evolution operator, it is straightforward to express the solution to the non-stationary Schrodinger equation in the following form: $$\phi(x,t) = \int \psi(x-x',t) \phi(x',0) dx'$$