# Why uncertainity is minimum for coherent states?

While reading for quantum damped harmonic oscillator, I came across coherent states, and I asked my prof about them and he said me it is the state at which $\Delta x\Delta y$ is minimum. I didn't quite understand why it is minimum.

• – Qmechanic Apr 10 '13 at 16:28

As you probably know, for any particle the product of the uncertainties in the position, $\Delta x$, and the momentum $\Delta p$ (not $\delta y$ as you state) is bounded below by a positive constant; $$\Delta x\, \Delta p\geq\frac{\hbar}{2}.$$ (If this doesn't ring a bell, you need to read up on Heisenberg's uncertainty principle.) For general states, the uncertainty product will probably be quite larger than $\hbar$, and for classical objects it will be very much larger. If we want to be very precise about a measurement, though, we would want the particle to have minimal uncertainty: that is, we'd want to impose the condition $$\Delta x\, \Delta p=\frac{\hbar}{2}.\tag{1}$$ The states that obey this condition are called coherent states.

A bit more technically, the general solutions to equation (1) are called squeezed coherent states, essentially because we can "squeeze" the uncertainty from $x$ into $p$ or vice versa. If the particle is in a harmonic-oscillator potential then we can choose a unique way to "split" the uncertainty product into "minimal" position and momentum parts, $$\Delta x=\sqrt{\frac{\hbar}{2m\omega}},\;\Delta p=\sqrt{\frac{1}{2}m\omega\hbar },$$ using the dimensional information contained in $\omega$ and $m$.

The states that satisfy $\Delta x\Delta p= \frac{\hbar}{2}$ (i.e. the strict equality rather than an inequality) are called "intelligent states". Coherent states are a subset of intelligent states, but not every intelligent state is coherent.

If you assume $\hat A$ and $\hat B$ are hermitian, and define, for a given state $\vert\psi\rangle$, the operator $$\Delta \hat A = \hat A - \langle A\rangle$$ then $\Delta \hat A$ is again hermitian. Defining now the shorthands $$\vert\psi_A\rangle = \Delta \hat A\vert\psi\rangle \, ,\qquad \vert\psi_B\rangle = \Delta \hat B\vert\psi\rangle$$ one uses the Schwarz inequality to show that $$\langle \psi_A\vert \psi_A\rangle \langle \psi_B\vert \psi_B\rangle \ge \vert \langle \psi_A\vert \psi_B\rangle\vert^2$$ For future reference, that $$\langle \psi_A \vert \psi_A \rangle\langle \psi_B \vert \psi_B \rangle= \vert\langle \psi_A \vert \psi_B \rangle\vert^2\Rightarrow \vert{\psi_A}\rangle =\mu\,\vert{\psi_B}\rangle\, , \tag{1}$$ i.e. the strict equality implies $\vert{\psi_A}\rangle$ is a scalar multiple of $\vert{\psi_B}\rangle$.

Expand \begin{eqnarray} \langle \psi_A \vert \psi_B \rangle&=& \langle \psi \vert\Delta \hat A\Delta \hat B\vert \psi \rangle \\ &=&\textstyle\frac{1}{2}\langle \psi \vert(\Delta \hat A\Delta \hat B-\Delta \hat B\Delta \hat A)\vert{\psi}\rangle +\textstyle{\frac{1}{2}}\langle {\psi}\vert (\Delta \hat A\Delta \hat B+\Delta \hat B\Delta \hat A)\vert \psi \rangle\, . \tag{2} \end{eqnarray} One easily shows that both terms on the right hand side of (2) are non-negative. If we keep only the $\textstyle\frac{1}{2}\langle \psi \vert(\Delta \hat A\Delta \hat B-\Delta \hat B\Delta \hat A)\vert{\psi}\rangle$ term we get the standard Robertson inequality, eventually written after some reorganization as
$$\Delta A\Delta B\ge \frac{1}{2}\vert \langle \psi\vert [\hat A,\hat B]\vert\psi \rangle\vert$$ To get the strict equality, we must additionally find states that satisfy Eq.(1) and simultaneously annul the second term in Eq.(2), i.e. states that simultaneous satisfy \begin{align} \Delta \hat A\vert\psi\rangle &=\mu \Delta \hat B\vert\psi\rangle\, , \tag{3} \\ \langle {\psi}\vert (\Delta \hat A\Delta \hat B+\Delta \hat B\Delta \hat A)\vert \psi \rangle&=0\, . \tag{4} \end{align} Using (3) and its complex conjugate one can rewrite (4) as $$0=\mu^*\langle{\psi}\vert (\Delta \hat B)^2\vert{\psi}\rangle +\mu \langle {\psi}\vert (\Delta\hat B)^2\vert{\psi}\rangle$$ but, since $\langle{\psi}\vert (\Delta \hat B)^2\vert{\psi}\rangle$ must be real, $\mu$ must be purely imaginary, i.e. $\mu=i \beta$, with $\beta$ real. Hence the condition on $\vert\psi\rangle$ is $$(\hat A-i\beta \hat B)\vert\psi\rangle =\lambda \vert\psi\rangle \, ,\qquad \lambda = \langle A\rangle -i \beta \langle B\rangle\, .$$ Moreover we also have $$(\Delta A)^2 =\beta^2 (\Delta B)^2 \tag{5}$$ Taking $\hat A=\hat x$ and $\hat B=\hat p$, and using $[\Delta \hat x,\Delta \hat p]=i\hat I$ one then shows that $$(\Delta p)^2 =-1/(2\beta)\, ,\qquad (\Delta x)^2=-\beta/2\, ,$$ implying that $\beta$ is negative and that $\beta=-\Delta x/\Delta p$. The intelligent states then satisfy \begin{align} (\hat x-i\beta \hat p)\psi(x)&= (x_0 -i\beta p_0)\psi(x)\, ,\\ &= (x+\beta \frac{d}{dx})\psi(x)\tag{6} \end{align} where $\langle x\rangle=x_0$ and $\langle p\rangle=p_0$.
The solution to (6) is (up to normalization) $$\psi(x)=C^{(x-\langle x\rangle)^2)/(2\beta) - i\langle p\rangle x} \tag{7}$$ The coherent state is the case for $\beta=-1$. With $\beta=-1$ the solution $\psi(x)$ is then just a Gaussian centred in $(x,p)$ at $(x_0,p_0)$. This is the coherent state, which by construction satisfies $\Delta x\Delta p=\frac{1}{2}\hbar$.

For all other values $-\infty<\beta <0$, with $\beta\ne -1$, the states are intelligent ($\Delta x\Delta p=\frac{1}{2}\hbar$ by construction) and squeezed since, by (5), the uncertainty in one of $x$ or $p$ is smaller than then uncertainty in the other.