[NOTES/QM-06011] Uncertainty Relation

The uncertainty, \(\Delta A)_\psi\), in a dynamical variable, \(X\) in a state, \(\psi\), is defined by
\begin{equation}
      (\Delta X)^2_\psi = \langle (\hat{X} -\overline{X})^2 \rangle_\psi .
\end{equation} Using this definition we derive an uncertainty relation between tow non commuting dynamical variables,.

Definition and properties of uncertainty $\Delta X$
Assuming  state vector $\ket{\psi}$ to be normalized, the average of a dynamical variable in the state $\ket{\psi}$is given by $\average{\psi}{X}$. The uncertainty of a physical quantity $X$ in a state $\psi$ will be denoted as $(\Delta X)_\psi$ and is defined by \begin{equation} (\Delta X)^2_\psi = \average{\psi}{\hat{X}^2} - \average{\psi}{\hat{X}} ^2, \label{EQ04} \end{equation} where notations $\overline{X}_\psi$ and $\langle X\rangle_\psi$ will be used to denote the average of an observable $X$ in state $\psi$ and omit the suffix $\psi$ to simplify intermediate steps.

Alternate Definition 1
The uncertainty $\Delta X$ can also be written in alternative form as \begin{equation} (\Delta X)^2_\psi = \langle (\hat{X} -\overline{X})^2 \rangle_\psi . \label{EQ05} \end{equation} We have

\[\begin{eqnarray} (\Delta X)^2_\psi  &=& \langle (\hat{X} -\overline{X})^2 \rangle_\psi \label{EQ06}\\ &=& \average{\psi}{(\hat{X}- \overline{X})^2}\label{EQ07} \\ &=& \average{\psi}{ \hat{X}^2 -\hat{X}\overline{X}-\overline{X}\hat{X} +   \overline{X}^2}\label{EQ08}\\   &=& \average{\psi}{\hat{X}^2 - 2\overline{X} \hat{X}  +\overline{X}^2}\label{EQ09}\\ &=& \average{\psi}{\hat{X}^2} -\average{\psi}{2\overline{X} \hat{X}}  +  \average{\psi}{\overline{X}^2} \label{EQ10}\\   &=&\average{\psi}{\hat{X}^2} - 2\overline{X}\average{\psi}{\hat{X}}        +\overline{X}^2\norm{\psi} =\overline{(X^2)} -        2\overline{X}^2 + \overline{X}^2\label{EQ11}\\&=& \overline{(X^2)} - \overline{X}^2\label{EQ12}\end{eqnarray}\]

Alternate Defintion 2
The above alternate definition can be cast in a useful form making use of the fact that $\hat{X}$ is hermitian and that its average valve $\bar{X}$ will always be real and hence
\begin{equation}
(\hat{X}-\bar{X})^\dagger=(\hat{X}^\dagger-\bar{X}^*)= (\hat{X}-\bar{X}).
\label{EQ13}
\end{equation}We now see that \begin{equation}
      (\Delta X)^2_\psi = \Norm{(\hat{X}-\bar{X})\psi}\label{EQ14}
\end{equation} Consider the right hand expression
\begin{eqnarray}
\Norm{(\hat{X}-\bar{X})\psi}^2
&=& \big((\hat{X}-\bar{X})\psi,\,
(\hat{X}-\bar{X})\psi\big)\\
&=& ( \psi,{(\hat{X}-\bar{X})^\dagger}(\hat{X}-\bar{X})\psi)\label{EQ15}\\
&=& ( \psi,(\hat{X}-\bar{X})^2\psi)        \label{EQ17}\\
&=& \langle(\hat{X}-\bar{X})^2 \rangle_\psi  \label{EQ18}
\end{eqnarray}

Expanding the expression in the right hand side of (\Ref{EQ05}), we get
\begin{eqnarray}
   (\Delta X)^2_\psi &=& \langle (X -\overline{X})^2 \rangle_\psi
                                                \label{EQ19}
\end{eqnarray}

Therefore, it follows that the uncertainty of a dynamical variable $X$ in a state $\ket{\psi}$ is zero  if and only if

\begin{eqnarray}
      (\hat{X}-\overline{X}) \ket{\psi} &=& 0 \label{EQ20}\\
       \text{or} \hat{X} \ket{\psi} &=& \overline{X} \ket{\psi}.\label{EQ21}
\end{eqnarray} This implies that the state $\ket{\psi}$ must be an eigenstate of the operator $\hat{X}$.

Generalized uncertainty relation

We will now prove the generalised uncertainty relation \EqRef{EQ03}. To simplify our notation we drop the suffix $\psi$ in expressions for uncertainties and averages. We also stop using $\hat{}$ for operators. Thus in the following  $A,B,C$ appearing in mathematical expressions will stand for the operators. As a first step we use the Cauchy Schwarz inequality \begin{equation} \Vert f\Vert^2 \Vert g\Vert^2 \ge |\innerproduct{f}{g}|^2 \Label{EQ22} \end{equation} with $\ket{f}=(A-\bar{A})\ket{\psi}$, and $\ket{g}=(B-\bar{B})\ket{\psi}$. So we have

\begin{eqnarray}
   \bra{f} &=& \bar{\psi}(A^\dagger-\bar{A}^*)=\bar{\psi}(A-\bar{A})
\label{EQ23}\\
   \bra{g} &=& \bar{\psi}(B^\dagger-\bar{B}^*) = \bar{\psi}(B-\bar{B})                                                                    \label{EQ24}
\end{eqnarray}  and therefore  \begin{eqnarray}
  \Norm{f}^2 = \norm{f} = \average{\psi}{(A-\bar{A})^2}          \label{EQ25}\\
  \Norm{g}^2 = \norm{g} = \average{\psi}{(B-\bar{B})^2}          \label{EQ26}\\
  \innerproduct{f}{g} = \average{\psi}{(A-\bar{A})(B-\bar{B})}     \label{EQ27}
 \end{eqnarray}inserting these expressions into Cauchy Schwartz inequality, we get
\begin{equation} (\Delta A)^2(\Delta B)^2 \ge |\average{\psi}{(A-\bar{A})(B-\bar{B}))}|^2\equiv |\average{\psi}{X}|^2, \label{EQ99} \end{equation}
where we have used $X$ to denote the operator expression $(A-\bar{A})(B-\bar{B}$. We write the operator $X$ as a sum of commutator and anticommutator of operators $(A-\bar{A})$ and $(B-\bar{B})$:
\begin{eqnarray}
   X = (A-\bar{A})(B-\bar{B})
   &=&\frac{1}{2} [(A-\bar{A}),(B-\bar{B})]_- +
\frac{1}{2}[(A-\bar{A}),(B-\bar{B})]_+ \\
 && \text{~~~~~~~~~ using  } XY = \frac{1}{2}[X,Y]_+ +\frac{1}{2} [X,Y]_- \\
   &=& \frac{1}{2}[(A-\bar{A}),(B-\bar{B})]_- +  \frac{i}{2} C.
\end{eqnarray}
This splitting of $X$ into two parts allows us to separate the real and
imaginary parts of $\average{\psi}{X}$ and to compute its absolute square
needed in \eqref{EQ99}:
\begin{equation}
   \average{\psi}{X}
   = \frac{1}{2}\underbrace{\average{\psi}{[(A-\bar{A}),(B-\bar{B})]_+}} +
\frac{i}{2}
\underbrace{\average{\psi}{C}} .
\end{equation}
Both the quantities marked with braces,being average values of hermitian operators, are {\it real}. Hence
\begin{equation}
   |\average{\psi}{X} |^2
   = \frac{1}{4}|\average{\psi}{[(A-\bar{A}),(B-\bar{B})]_+}|^2 +
          \frac{1}{4} |\average{\psi}{C}|^2 .
\end{equation}
Therefore the inequality \rqref{EQ99} becomes
\begin{eqnarray}
    (\Delta A)^2(\Delta B)^2 &\ge& \frac{1}{4}
|\underline{\average{\psi}{[(A-\bar{A}),(B-\bar{B})]_+}}|^2 + \frac{1}{4}
|\underline{\average{\psi}{C}}|^2  \Label{EQ97}\\
\therefore \quad (\Delta A)^2(\Delta B)^2 &\ge& \frac{1}{4}
|\average{\psi}{C}|^2.
\end{eqnarray}

This gives us the desired generalised form of uncertainty relation \begin{equation} \Delta A \Delta B \ge \frac{1}{2} \langle C \rangle_\psi. \Label{EQ98} \end{equation} For position and  momentum $[\hat{x},\hat{p}]=-\hbar $, and hence we have \begin{equation} (\Delta x)_\psi\, (\Delta p)_\psi \ge \frac{\hbar}{2}.\average{\psi}{C} \end{equation}

Minimum uncertainty states
{\sf Let us now ask,``Which states  will have the minimum value of the uncertainty product?''} Going back to the first step and analysing the proof, for equality to hold in the uncertainty relation \eqref{\TheFile;EQ98}, the Cauchy Schwarz inequality, \eqref{EQ22}, must become equality. This will be the case if and only if $\ket{f} = \lambda ket{g}$, for some complex number $\lambda$, In addition to this the average of the anticommutator in \eqref{EQ97} is zero. Thus the condition for state $\ket{\psi}$ to be state with minimum uncertainty becomes

\begin{eqnarray}
 (A-\bar{A})\ket{\psi} =\lambda (B-\bar{B}) \ket{\psi},\qquad
 [(A-\bar{A}),(B-\bar{B})]_+\ket{\psi} =0.
\end{eqnarray}