Uncertainty Principle
- Quantum Mechanics 1.8
from 「Modern Quantum Mechanics」: Sakurai, J. J.
The fundamental limit of measurements.
- Standard deviation of a measurement
- Robertson–Schrödinger uncertainty relations
- Heisenberg limit
- Simultaneously measurable quantities
Standard deviation of a measurement
For an arbitrary Hermitian operator $\hat{L}$ we can associate a standard deviation:
\[ \begin{align*} \sigma_L^2 &=\expct{\hat{L}^2}-\expct{\hat{L}}^2 \nl &= \expct{\hat{L}^2}-2\expct{\hat{L}}\expct{\hat{L}}+\expct{\hat{L}}^2 \nl &= \Brktop{\psi}{\hat{L}^2-2\hat{L}\expct{\hat{L}}+\expct{\hat{L}}^2}{\psi} \nl &= \Brktop{\psi}{(\hat{L}-\expct{\hat{L}})^2}{\psi} \nl &= \Brkt{(\hat{L}-\expct{\hat{L}})\psi}{(\hat{L}-\expct{\hat{L}})\psi} \end{align*} \]
Using $\hat{\delta}_L=\hat{L}-\expct{\hat{L}}$, we can write it simply.
\[ \sigma_L^2=\Brkt{\hat{\delta}_L\psi}{\hat{\delta}_L\psi} \]
Robertson–Schrödinger uncertainty relations
The product of the two deviations of Hermitian operators $\hat{A}$ and $\hat{B}$ can thus be expressed as
\[ \sigma_A^2\sigma_B^2 = \Brkt{\hat{\delta}_A\psi}{\hat{\delta}_A\psi}\Brkt{\hat{\delta}_B\psi}{\hat{\delta}_B\psi} \]
By the Cauchy–Schwarz inequality,
\[ \begin{align*} \sigma_A^2\sigma_B^2 &\ge \abs{\Brkt{\hat{\delta}_A\psi}{\hat{\delta}_B\psi}}^2 \nl &= \abs{\expct{ \hat{\delta}_A\hat{\delta}_B }}^2 \nl &= \Re\left( \expct{\hat{\delta}_A\hat{\delta}_B} \right)^2 + \Im\left( \expct{\hat{\delta}_A\hat{\delta}_B} \right)^2 \nl &= \abs{\frac{1}{2}\left[ \expct{\hat{\delta}_A\hat{\delta}_B}+\expct{\hat{\delta}_B\hat{\delta}_A} \right]}^2 + \abs{\frac{1}{2i}\left[ \expct{\hat{\delta}_A\hat{\delta}_B}-\expct{\hat{\delta}_B\hat{\delta}_A} \right]}^2 \end{align*} \]
We find that
\[ \begin{align*} \expct{\hat{\delta}_A\hat{\delta}_B} &= \Expct{ (\hat{A}-\expct{\hat{A}})(\hat{B}-\expct{\hat{B}}) } \nl &= \Expct{ \hat{A}\hat{B}-\hat{A}\expct{\hat{B}}-\hat{B}\expct{\hat{A}}+\expct{\hat{A}}\expct{\hat{B}} } \nl &= \expct{\hat{A}\hat{B}}-\expct{\hat{A}}\expct{\hat{B}} \end{align*} \]
We now substitute this back into the equation above and get
\[ \begin{align*} \sigma_A^2\sigma_B^2 &\ge \abs{\frac{1}{2}\left[ \expct{\hat{A}\hat{B}}+\expct{\hat{B}\hat{A}}-2\expct{\hat{A}}\expct{\hat{B}} \right]}^2 + \abs{\frac{1}{2}\left[ \expct{\hat{A}\hat{B}}-\expct{\hat{B}\hat{A}} \right]}^2 \nl &= \abs{\frac{1}{2}\left[ \expct{\acomm{\hat{A}}{\hat{B}}}-2\expct{\hat{A}}\expct{\hat{B}} \right]}^2 + \abs{\frac{1}{2}\left[ \expct{\comm{\hat{A}}{\hat{B}}} \right]}^2 \end{align*} \]
Then we finally get the Schrödinger uncertainty relation: \[ \sigma_A\sigma_B \ge \frac{1}{2}\sqrt{ \abs{\expct{\acomm{\hat{A}}{\hat{B}}}-2\expct{\hat{A}}\expct{\hat{B}}}^2 + \abs{\expct{\comm{\hat{A}}{\hat{B}}}}^2 } \]
There is also the Robertson uncertainty relation that has a slightly weaker condition, a more familiar form. \[ \sigma_A\sigma_B \ge \frac{1}{2}\abs{\expct{\comm{\hat{A}}{\hat{B}}}} \]
Example
Let’s find uncertainty about the measurement of position and momentum.
Since $\comm{\hat{x}}{\hat{p}}=i\hbar$,
- $ \sigma_x\sigma_p \ge \dfrac{\hbar}{2} $
Heisenberg limit
In quantum metrology, and especially interferometry, the Heisenberg limit is the optimal rate at which the accuracy of a measurement can scale with the energy used in the measurement. Typically, this is the measurement of a phase (applied to one arm of a beam-splitter) and the energy is given by the number of photons used in an interferometer.
Simultaneously measurable quantities
By the (Robertson) uncertainty relation, there exists a Heisenberg limit of two physical quantities if the two respective operators are not commutating. However, if the two operators commute, then there isn’t a Heisenberg limit, which means the two quantities are simultaneously measurable.
