Extension of uncertainty principle to mixed-states

Question

In his Lectures on Quantum Theory, Isham states the generalized uncertainty principle for two observables $O_1$ and $O_2$ for a pure state $|\psi\rangle$ as $$\Delta_\psi O_1~\Delta_\psi O_2 \ge {1\over 2}\Bigl|\langle\psi|[O_1,O_2]|\psi\rangle\Bigr|.$$

Question: To generalize this to a mixed-state with density matrix $\hat{\rho}:=\sum_{d=1}^Dw_d|\psi_d\rangle\langle\psi_d|$, is it appropriate to write $\Delta_\rho O_1=\sum_{d=1}^Dw_d\Delta_{\psi_d}O_1$?

Also, if I plug in this expression in the above generalized expression, then I get terms like $\Delta_{\psi_i} O_1 ~\Delta_{\psi_j}O_2$ in the expression for $\Delta_\rho O_1~\Delta_\rho O_2$. What to do when $i\ne j$?

Answer 1

The procedure would be the same as the procedure for finding the uncertainty principle for pure states. We define $\Delta O$ as $\sqrt{\langle O^2\rangle- \langle O \rangle^2 }$. Since the two expectation values that go into calculating the uncertainty can be calculated from the density matrix via the usual expression for the expectation value of an operator $\langle O\rangle = \text{Tr}(O\rho)$.

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The expression for $\Delta O$ as $\sqrt{\langle O^2\rangle- \langle O \rangle^2 }$ is so hostile to linearity that we cannot hope for a simple linear expression of the kind you propose for the uncertainty of an observable. This has been explicitly demonstrated in the answer by @user1723984. However, the Robertson-Schrodinger uncertainty relation for the product of the uncertainties of two observables has a nicer expression for a pure state $\psi$, in particular, \begin{align}\sigma^2_A\sigma^2_B \ge {1\over 4}\Bigl|\langle\psi|[A,B]|\psi\rangle\Bigr|^2\end{align} as you notice. Let's see if we can recover an analog of this relation for the case of the density matrix. We write

\begin{align} \sigma^2_A&=\Bigr(\text{Tr}\big(\rho A^2\big)-\text{Tr}\big(\rho A\big)\text{Tr}\big(\rho A\big)\Bigr)\\ &=\sum_i\bigg\langle i\bigg|\rho A^2-\sum_j\Big\langle j\Big|\rho A\Big|j\Big\rangle \rho A\bigg|i\bigg\rangle\\ &=\sum_i\bigg\langle i\bigg|\rho A\Big(A-\sum_j\Big\langle j\Big|\rho A\Big|j\Big\rangle \Big) \bigg|i\bigg\rangle\\ &=\text{Tr}\Big(\rho A\big(A-\langle A\rangle\big)\Big)\\ &=\text{Tr}\Big(\rho\big(A-\langle A\rangle\big)^2\Big)\\ \end{align} Thus, \begin{align} \sigma^2_A\sigma^2_B&=\text{Tr}\Big(\rho\big(A-\langle A\rangle\big)^2\Big)\text{Tr}\Big(\rho\big(B-\langle B\rangle\big)^2\Big)\\ \end{align} Now, the Cauchy-Schwarz inequalities hold true for all inner products, and it can be shown that for Hermitian operators $X$ and $Y$, it takes the following form

\begin{align} \big|\text{Tr}\big(SX Y\big)\big|^2 \leq \text{Tr}\big(SX^2\big)\text{Tr}\big(SY^2\big) \end{align} where $S$ is a semi-positive definite matrix. Since $\rho$ is a semi-positive definite matrix and $A-\langle A\rangle$, and $B-\langle B\rangle$ are Hermitian operators, an application of this inequality yields \begin{align} \sigma_A^2\sigma_B^2&\geq \bigg|\text{Tr}\Big(\rho \big(A-\langle A\rangle\big)\big(B-\langle B\rangle\big)\Big)\bigg|^2\\ &= \Big|\text{Tr}( \rho AB ) - \text{Tr} (\rho A)\text{Tr} (\rho B)\Big|^2 \end{align}

Exploiting the facts that

$1.$ modulus square should be bigger than the square of its imaginary part,

$2.$ $\overline{\text{Tr}(\rho A B)}=\text{Tr}(\rho BA)$, and

$3.$ $\overline{\text{Tr}(\rho A)}=\text{Tr}(\rho A)$, we can write

\begin{align} \sigma_A^2\sigma_B^2&\geq\frac{1}{4}\Big|{\text{Tr}(\rho AB)-\text{Tr}(\rho BA)}\Big|^2\\ \implies\sigma_A\sigma_B&\geq\frac{1}{2}\Big|\text{Tr}(\rho [A,B])\Big| \end{align} which is the exact equivalent of the Robertson-Schrodinger uncertainty relation. Now, expanding the density matrix as $\rho = \sum_n p_n|\psi_n\rangle\langle \psi_n |$, we can easily see that $$\text{Tr}(\rho [A,B])=\sum_np_n\langle\psi_n|[A,B]|\psi_n\rangle$$

Thus, even tho not for individual uncertainties, for the uncertainty principle, we can write down

$$\Delta_\rho A\Delta_\rho B \geq \sum_n p_n \Delta_{\psi_n} A \Delta_{\psi_n} B$$

where I've switched the notation in a self-explanatory manner.

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I might have made some sign/modulus mistakes, so kindly verify.

Answer 2

No, the standard deviation is not linear in the state

$$ \Delta_\psi O=\sqrt{\langle O^2\rangle_\psi-\langle O\rangle_\psi^2}\tag{1}$$

$$\Delta_{\alpha \psi + \beta\phi} O=\sqrt{\langle O^2\rangle_{\alpha \psi + \beta\phi}-\langle O\rangle_{\alpha \psi + \beta\phi}^2} =\sqrt{\langle O^2\rangle_{\alpha \psi}+\langle O^2\rangle_{\beta \phi}+ (\langle O\rangle_{\alpha \psi}+\langle O\rangle_{\beta \phi})^2}\neq\\\neq\alpha\Delta_\psi O+\beta\Delta_\phi O$$

Instead you can simply use $(1)$ and

$$ \langle O\rangle_\rho=\mathrm{Tr}(\rho O)$$

Answer 3

Consider the variance of an observable $A$ with respect to a generic mixed state $\rho$: $$\operatorname{Var}[A|\rho] \equiv \operatorname{tr}(\rho A^2)-\operatorname{tr}(\rho A)^2,$$ also written concisely as $\sigma_A^2\equiv\operatorname{Var}[A|\rho]$, $\langle A\rangle\equiv \operatorname{tr}(\rho A)$, and thus $\sigma_A^2=\langle A^2\rangle-\langle A\rangle^2$. Yet another formulation for generic variances is $$\operatorname{Var}[A|\rho] = \operatorname{tr}[\rho (A- \langle A\rangle)^2].$$ Let's now define the following inner product in the subspace of Hermitian operators: $$\langle A,B\rangle_\rho \equiv \operatorname{tr}[A \rho B] = \operatorname{tr}[\rho BA].$$ You can directly verify that this is indeed an inner product, as long as $\rho$ is positive semidefinite (which it is, if it's a state).

With respect to this inner product we can now write concisely $$\operatorname{Var}[A|\rho] = \|A-\langle A\rangle_\rho\|^2_\rho, \quad \|C\|_\rho^2\equiv\langle C,C\rangle_\rho,$$ and $\langle A\rangle_\rho\equiv\operatorname{tr}(A\rho) = \langle A,I\rangle_\rho$.

We can now finally apply the CS inequality with respect to this $\rho$-inner product structure, obtaining $$\operatorname{Var}[A|\rho]\operatorname{Var}[B|\rho] \ge \lvert \langle A- \langle A\rangle_\rho,B-\langle B\rangle_\rho\rangle_\rho\rvert^2 \\ = \lvert \langle A,B\rangle_\rho - \langle A\rangle_\rho \langle B\rangle_\rho \rvert^2.$$ This is precisely the uncertainty principle in its generalised form. Taking real and imaginary components of the modulus and neglecting the real component you get the standard formulation involving only the commutator $[A,B]$.