Quantum State Function $\psi$

Question

If I write in QM at an instant, that the quantum state that describes the particle completely at an instant $\psi(x)=\cos(6\pi x)$. Does that mean $|\psi(x)|^2dx$ after normalisation gives me the probability that particle will be positioned between $x$ and $dx$ ? What if instead of position it was a function for state of momentum/energy, could I apply born's rule for getting wave function for position ?

Answer 1

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. The probability that $x$ will be in the interval $[a, b]$ is then $$ P_{a\le x\le b} (t) = \int\limits_a^b d x\,|\psi(x)|^2 \, . $$ Normalization of $\psi(x)$ to one is required since the probabilities of all possible outcomes should add up to one (the particle is certainly somewhere) $$ \int\limits_{-\infty}^\infty |\psi(x)|^2 \mathrm{d}x = 1 \, . $$

If you work with momentum wave function $\psi(p)$, then $|\psi(p)|^2 \mathrm{d}p$ gives you the probability that momentum of the particle is between $p$ and $p + \mathrm{d}p$.

The momentum wave function is related to the position one by a Fourier transform $$ \psi(x)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(p) e^{\mathrm{i} p x / \hbar} \mathrm{d}p \, , $$ while the position wave function is related to the momentum one by an inverse Fourier transform $$ \psi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(x) e^{-\mathrm{i} p x / \hbar} \mathrm{d}x \, . $$