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The probability amplitude for a free particle with momentum $\ p$ and energy $E$ is the complex wave function: $$\psi_{(\vec x , t)}=e^{i(\vec k\cdot \vec x -\omega t)}$$ is there any uncertainty on the free particle with a definite momentum $\vec p$!? |
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The question is a bit more subtle than you (probably) think. The wavefunction you've given is for an infinite plane wave with a perfectly defined momentum. Because the wave is infinite the probabilty of finding the particle is the same everywhere, i.e. the uncertainty in the particle position is infinite. This is the same result you get by setting $\sigma_p$ to zero in the uncertainty relation dmckee gave in his answer. If you want a wavefunction for a particle with a position you can can build one by adding up a number of plane waves with different values of momentum, $p$. The example usually given in QM courses is to make the particle wavefunction a gaussian curve. You may think it's hard to make a gaussian by adding up plane waves, but actually it's very easy. There's a mathematical procedure called a Fourier transform that you can use to disassemble a gaussian into it's component plane waves. If you're interested, this calculation is given in detail here. With a gaussian we normally describe it's width by the width at 1/e of the peak height, and this width is known as $\sigma$. If you use a gaussian with a width $\sigma_x$ to describe your particle, then the Fourier transform gives a distribtuion of momenta that is a gaussian of width $\sigma_p$, and as the calculation I linked to shows, $\sigma_x$ and $\sigma_p$ are related by: $$\sigma_p \sigma_x = \frac{\hbar}{2}$$ exactly as the uncertainty relation requires. |
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Yes there is. If you attempt to compute the expectation value of the particle's position at a particular time you are going to run into big trouble. You'll find it has a infinite range of possible positions. $$ \sigma_p \sigma_x \ge \frac{\hbar}{2} $$ If you send $\sigma_p$ toward zero... |
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