# Is it only the ground state of the quantum harmonic oscillator that has the minimum uncertainty product?

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?

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Gaussian wavepacket has the minimum uncertainty product . Evidently the ground state eigenfunction of QHO is Gaussian. Secondly for a general eigenstate of QHO : $\Delta x \Delta p = (n + \frac{1}{2})h$ – user35952 Feb 5 '14 at 5:41
Related: physics.stackexchange.com/q/60655/2451 and links therein. – Qmechanic Feb 5 '14 at 8:24
@Qmechanic: So the general answer is "squeezed coherent states", while for the balanced situation (when we also require $\Delta p =\Delta x$) the answer is "coherent states". – Zoltan Zimboras Feb 5 '14 at 8:28

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.

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I agree with Lubos that the $n$th excited state of the harmonic oscillator Hamiltonian $H= x^2 + p^2$ does not minimize $\Delta x \Delta p$, only its ground state does. However, I have to disagree with him a bit that only states of the form

$$\psi(x) = A \cdot \exp \left( -a x^2 + b \right)$$

minimize this. Let $\psi_n(x)$ denote the $n$th excited state of the Harmonic oscillator (with $\psi_0$ being the ground state), then for any $\alpha$ nonzero complex parameter define the state

$$\phi_\alpha (x)= e^{-\tfrac{1}{2}|\alpha|} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \psi_n (x).$$

These are the so-called coherent states (they are states of form $A \cdot e^{-a(x-d)^2 + icx}$ - with $a,c,d$ depending on $\alpha$) and they also minimize $\Delta x \Delta p$. In fact they are the only states (together with $\psi_0$) that minimize this product in a balanced way i.e. with $\Delta x =\Delta p$.

Edit. The links in Qmechanic's link (see comment section) are very useful, and answer fully the question: the most general states that minimize $\Delta x \Delta p$, without the requiring $\Delta x =\Delta p$, are the squeezed coherent states.

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$a$ and $b$ could be complex in Lubos's answer. – Nogueira Feb 27 '15 at 13:26