Return to Answer

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. Normalisation (to one) is not required actually, it just desirable to have probabilities in the conventional range $[0,1]$.

This works, of course, if $\psi(x)$ is the wave function for a system, and not just an arbitrary function of $x$.

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 $$ \psi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(x) e^{-\mathrm{i} p x / \hbar} \mathrm{d}x \, . $$

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 $$ \psi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(x) e^{-\mathrm{i} p x / \hbar} \mathrm{d}x \, . $$

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. Normalisation (to one) is not required actually, it just desirable to have probabilities in the conventional range $[0,1]$.

This works, of course, if $\psi(x)$ is the wave function for a system, and not just an arbitrary function of $x$.

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 $p + \mathrm{d}p$.

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. Normalisation (to one) is not required actually, it just desirable to have probabilities in the conventional range $[0,1]$.

This works, of course, if $\psi(x)$ is the wave function for a system, and not just an arbitrary function of $x$.

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 $$ \psi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(x) e^{-\mathrm{i} p x / \hbar} \mathrm{d}x \, . $$

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. Normalisation (to one) is not required actually, it just desirable to have probabilities in the conventional range $[0,1]$.

This works, of course, if $\psi(x)$ is the wave function for a system, and not just an arbitrary function of $x$.

Yes, $|\psi(x)|^2 \mathrm{d}x$ gives you the probability to find the particle between $x$ and $x + \mathrm{d}x$. Normalisation (to one) is not required actually, it just desirable to have probabilities in the conventional range $[0,1]$.

This works, of course, if $\psi(x)$ is the wave function for a system, and not just an arbitrary function of $x$.

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 $p + \mathrm{d}p$.