We know that the uncertainty product of general states is bounded by the inequality described by Heisenberg's uncertainty relation. And the ground state of the quantum harmonic oscillator falls under the minimum uncertainty product ($\frac{\hbar}{2}$). Do the excited states of QHO have also the minimum uncertainty product or higher?
[Physics] Is it only the ground state of the quantum harmonic oscillator that has the minimum uncertainty product
harmonic-oscillatorheisenberg-uncertainty-principlequantum mechanics
Best Answer
Excited states of the harmonic oscillators don't minimize the product: both $\Delta x$ and $\Delta p$ scale like $\sqrt{E}\sim \sqrt{N+1/2}$ in the $N$-th excited state (just realize that $H$ is a combination of $p^2$ and $x^2$ whose expectation values we are computing to get $\Delta x$ and $\Delta p$) so it is only minimized for $N=0$.
The most general wave function that minimizes the product is a Gaussian $$ \psi(x) = A\cdot \exp(-ax^2+b) $$ for some constants $A,a,b$ where $a$ has to be real, I guess. This generalizes the ground state of the harmonic oscillator by a "squeezing" (given by the constant $a$) in the $x$-space that is equivalent to the "expanding" in the $p$-space; by translation in the $x$-direction given by the real part of $b$; and translation in the $p$ direction given by the imaginary part of $b$. The normalization constant $A$ is arbitrary – or has to be calculated if you want a wave function normalized to one.