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A quantum linear harmonic oscillator has a definite non-zero ground state energy $E_0=\frac{1}{2}\hbar\omega\neq 0$. However, in this energy eigenstate, the position and momenta are uncertain and their standard deviations satisfy an uncertainty relation $(\Delta x\Delta p_x)_{|0\rangle}=\frac{\hbar}{2}$. I want to ask whether this uncertainty is related to the fact that the ground state energy is nonzero and if yes, how exactly this value of $E_0$ is obtained?

Can I extrapolate this inference in free quantum field theory (such as free Klein-Gordon theory)? A free KG field has an infinite ground state energy. Can it be attributed to an uncertainty relation acting between the field $\phi(x)$ and the corresponding conjugate momentum operator $\pi(x)$?

SRS
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It can easily be seen that there is no direct relationship between the zero point energy and the uncertainty relation. If the Hamiltonian is

$$H = \frac{p^2}{2m} + \frac12 m \omega^2 x^2$$

then defining

$$H' = H - \frac12 \hbar \omega$$

the zero point energy disappears but the uncertainty remains, since adding a constant to the Hamiltonian won't change any observable (as long as we steer clear of gravity).

This is what is usually done in QFT: most books say something along the lines of "the zero point energy is infinite, so let's substract this infinite contribution from the Hamiltonian and be done with it". The commutation relations aren't affected, because they are independent from the Hamiltonian.

Javier
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