# Why don't cancelling wavefunctions for two different particles give zero total wavefunction?

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?

• 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

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.