# Uncertainty relation for non-simultaneous observation

Heisenberg's uncertainty relation in the Robertson-Schroedinger formulation is written as,

$$\sigma_A^2 \sigma_B^2 \geq \left|\frac{1}{2} \langle\{\hat A, \hat B\}\rangle -\langle \hat A\rangle\langle \hat B\rangle\right|^2+\left|\frac{1}{2}\langle[\hat A,\hat B]\rangle\right|^2$$ where $$\sigma_A^2 = \langle\psi|(\hat A-\langle \hat A \rangle)^2 |\psi\rangle$$ and $$\sigma_B^2 = \langle\psi|(\hat B-\langle \hat B \rangle)^2 |\psi\rangle$$ calculated in the same state $$\psi$$ for both observables $$\hat A$$ and $$\hat B$$.

Now my question is what happens to the other side of the inequality if we calculate one variance for state $$\psi(t)$$ and then let the state evolve to $$\psi(t+\delta t)$$ and now calculate the other variance in the product. In other words, what is the QM lower limit of this product: $$\langle {\psi(t)|(\hat A -\langle \hat A\rangle)^2|\psi(t)\rangle} ~\langle {\psi(t+\delta t)|(\hat B -\langle \hat B\rangle)^2|\psi(t+\delta t)\rangle}$$ for arbitrary $$\delta t$$ and $$\psi(t)$$ is evolving according to the time-dependent Schroedinger equation $$\hat H \psi(t)=i \hbar\frac{\partial \psi(t)}{\partial t}~?$$

• I am surprised that this is construed as a "homework" or "off topic" problem. I have never seen asked before. Is the answer that obvious to you @Danu, Prahar, Jim, JamalS, Brandon Enright? The "RHS" must somehow depend on $\hat H$ I just don't know how, and down-voting does not solve the problem. The answer offered below is obviously and admittedly wrong. – hyportnex Nov 12 '14 at 0:15
• Since $|\psi(t)\rangle=\exp\left(it\hat{H}/\hbar\right)|\psi(0)\rangle$, you probably could write $|\psi(t+\delta t)\rangle$ using this time-evolution operator. This probably would require you to know $[\hat{H},\,\hat{B}]$. – Kyle Kanos Nov 18 '14 at 3:37
• In short, this can be zero. In the Heisenberg picture, it is possible for $B(0)$ to be canonically conjugate to $A(0)$, and then 'rotate' into $B(\Delta t)=A(0)$. Examples are easy to find with $x$ and $p$ in a harmonic oscillator, or $\sigma_x$ and $\sigma_y$ for a two-level system where $H\propto \sigma_z$. – Emilio Pisanty Nov 18 '14 at 12:20
• Physical note: There are no "non-simultaneous observations" here. Either you measure a system in which case the quantum state will be altered and after some time you'll NOT have $|\psi(t+\delta t)\rangle$, or it won't, in which case you don't have an observation. Also note that the Robertson-Schrödinger uncertainty relation is not about simultaneous measurement, it is about variances of observables after preparation. – Martin Sep 24 '15 at 16:54
• Technically, no, it is not an observarble, because of the state-dependent part $\langle A \rangle$, and the notation $\langle A \rangle$ might be ambiguous.in this case. – pppqqq Dec 5 '16 at 20:33

## 1 Answer

Try to use the Heisenberg picture. It will become more evident what you are actually computing, namely the uncertainty relation between two obsevables $$A(t_1)$$ and $$B(t_2)$$. You can take as an exercise the harmonic oscillator and compute the uncertainty relation between $$x(0)$$ and $$x(t)$$, it will be nonzero for $$t$$ different from the period of the oscillator.

hint:

$$x(t)=e^{iHt}x(0)e^{-iHt}$$