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Let $\left|a\right>=e^{i(kx-\omega t)}$, $\left|b\right>=-e^{i(kx-\omega t)}$ be two neutral particles in the 1D free space without any interaction. Then $E_a=\left<a\left|\hat{H}\right|a\right>=\hbar\omega$, $E_b=\left<b\left|\hat{H}\right|b\right>=\hbar\omega$. Hence $E_a+E_b=2\hbar \omega$. That means the energy of the system $\{a,b\}$ is $2\hbar \omega$.

However, in another way, $\left|a\right>+\left|b\right>=0$, hence $E_{a+b}=(\left<a\right|+\left<b\right|)\hat H(\left|a\right>+\left|b\right>)=0$. That means the total energy is $0$.

These two method seems both make some sense, but they are contradict to each other. What's wrong with them?

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The total energy for the zero wavefunction is undefined, because it is not normalized. The fallacy is adding wavefunctions for different particles--- wavefunctions give amplitudes for configurations of the whole universe (or for the whole "isolated system"), not for particles one at a time. –  Ron Maimon Jul 6 '12 at 8:29
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up vote 4 down vote accepted

The wavefunction of two particles is not the sum of the wavefunction for the individual particles. The total wavefunction is the product of the two wavefunctions for finding particle 1 at $x_1$ at time t and finding particle 2 at $x_2$ at time t:

$$ \psi(x_1,x_2,t) = e^{i(kx_1 + k x_2 - 2\omega t)} $$

The energy is $2\omega$. The fallacy of adding wavefunctions for different particles is common. When you have a bigger system, the Hilbert space is the tensor product of the two spaces, and indepdendent single particle wavefunctions are product vectors inside the tensor product.

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I see, thank you. –  Popopo Jul 6 '12 at 8:31
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