# How do you describe a quantum state vector mathematically?

Multiple sources I have used have said that a quantum state vector $|x\rangle$ can be taken to be or approximated to be:

$$|ψ\rangle = \mathrm e^{-\omega x^2/2}$$

Where $\omega$ is the angular freequency. I would just like an explanation for why this is and how it came about.

The first section it's defining the wavefunction ψ(x), and it gives an equation with reference Hermite polynomials and spectral method etc... Assuming a mass, m, of 1 and by temporarily removing the reduced-Planck constant we take the middle term which was, according to Wikipedia: $$|ψ\rangle = \mathrm e^{-\omega mx^2/2ħ }$$Under the new conditions to be as stated before: $$|ψ\rangle = \mathrm e^{-\omega x^2/2}$$

• Friendly reminder: always cite your sources. Commented May 15, 2017 at 16:30
• I don't think this a super standard thing. Usually states are defined in context with their Hamiltonian. For example, you'll often see $|x \rangle$ to mean a complete set of states such that $\langle x | \psi \rangle = |\psi(x) \rangle$. $|x\rangle$ certainly could not mean $e^{wx^2}$ in that case. Commented May 15, 2017 at 16:43
• The (incorrect, and ultimately meaningless) formula you quote is not in the source you provide. $-1$. Commented May 15, 2017 at 21:12

In practice, the quantum state for a particular problem is most often expressed with projections on a convenient basis of the Hilbert space of all quantum state. Typically useful one are $|x\rangle$ (beware, not the same as yours: I use standard notations here contrary to you), which are the eigenvectors of the position operator, or $|p\rangle$, which are the eigenvectors of the momentum operator. Now given a ket $|\psi\rangle$, the projection $\langle x|\psi\rangle$ is just what is usually referred to as the wave function, the one appearing in Schrödinger equation. This is what is done in that Wikipedia article you quote. The dependence on the position exhibited in that case, $\langle x|\psi\rangle \propto \frac{m\omega x^2}{2\hbar}$, is a consequence of the potential used in said Schödinger equation. Note that one could also work with $\langle p|\psi\rangle$ in that example, which would be proportional to $\exp\frac{-p^2}{2\hbar m \omega}$.