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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 ?

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  • $\begingroup$ Since this is tagged homework, I'll ask you to take one more step: normalize the wavefunction, instead of saying "after normalization". $\endgroup$ – garyp Mar 5 '14 at 12:29
  • $\begingroup$ @garyp done :) . $\endgroup$ – Isomorphic Mar 5 '14 at 12:30
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    $\begingroup$ @lota You apply a Fourier transform to get the position wave function out of the momentum one, not the Born rule. $\endgroup$ – Wildcat Mar 5 '14 at 12:53
  • $\begingroup$ The function is still not normalized. $\endgroup$ – garyp Mar 5 '14 at 12:58
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    $\begingroup$ @Wildcat it could be a wavefunction for infinite potential well, $x\in[-\frac14,\frac14]$, for example. There it'd be square integrable. $\endgroup$ – Ruslan Mar 5 '14 at 16:57
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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 $$ \psi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi(x) e^{-\mathrm{i} p x / \hbar} \mathrm{d}x \, . $$

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  • $\begingroup$ Thank you, I have edited the question slightly.Please see. $\endgroup$ – Isomorphic Mar 5 '14 at 12:29
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    $\begingroup$ The range is $[0,1]$ is more than desirable or conventional, it is the first axiom of probability. $\endgroup$ – innisfree Mar 5 '14 at 12:56
  • $\begingroup$ @Iota What is the meaning of a probability outside of the range [0,1]? $\endgroup$ – garyp Mar 5 '14 at 12:59
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    $\begingroup$ @Wildcat that % symbol looks like and basically means divide by 100. Percentages are normalized to 1. You can't pick any other normalization for a probability. That is literally the first axiom of probability. If you do, you are no longer talking about a probability. $\endgroup$ – innisfree Mar 5 '14 at 14:06
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    $\begingroup$ Normalization in fact could indeed be not necessary. One could e.g. be interested in ratio of probabilities between two points in spacetime, then the normalization coefficients would just cancel out. $\endgroup$ – Ruslan Mar 5 '14 at 16:59

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