Heisenberg Uncertainty Relation for mixed states

[Jamister]

TL;DR Summary

proving the Heisenberg uncertainty relations for mixed states

How do you prove Heisenberg uncertainty relations for mixed states (density matrix), only from knowing the relation is true for pure states?

 

[hilbert2]

I wouldn't expect the uncertainty in any observable to get smaller by adding statistical uncertainty to the already existing quantum uncertainty in the position and momentum. It can only increase.

There's something said about this on page 16 here: https://www.univie.ac.at/nuhag-php/dateien/talks/1073_NuHAG_WPI_2008.pdf

 

[vanhees71]

You can prove it for any state. Let ##\hat{A}## and ##\hat{B}## be the self-adjoint operators representing observables and assume for simplicity that ##\langle A \rangle=\langle B \rangle=0##, where ##\langle \hat{A} \rangle=\mathrm{Tr}(\hat \rho \hat{A})##. Since ##\hat{\rho}## is a positive semi-definite self-adjoint operator there's a complete orthonormalized eigenbasis ##|u_n \rangle##, i.e.,
$$\hat{\rho}=\sum_n p_n |u_n \rangle \langle u_n |.$$
Thus any expectation value reads
$$\langle A \rangle = \mathrm{Tr} (\hat{\rho} \hat{A}) = \sum_{n} \langle u_n| \rho_n \hat{A} |u_n \rangle = \sum_n P_n \langle u_n |\hat{A}| u_n \rangle.$$
Now apply this to the self-adjoint operator
$$\hat{C}=(\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B}), \quad \lambda \in \mathbb{R}.$$
First of all
$$\langle u_n|\hat{C}| u_n \rangle=\langle (\hat{A}+\mathrm{i} \lambda \hat{B}) u_n|(\hat{A}+\mathrm{i} \lambda \hat{B}) u_n \rangle \geq 0.$$
And since ##P_n \geq 0## for all ##n## you also have
$$P(\lambda)=\langle (\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B}) \rangle \geq 0.$$
Multiplying this out gives
$$P(\lambda) = \langle A^2 \rangle + \langle \mathrm{i} [\hat{A},\hat{B}] \rangle \lambda + \lambda^2 \langle \hat{B}^2 \rangle.$$
Now ##\langle A^2 \rangle=\Delta A^2## and ##\langle B^2 \rangle=\Delta B^2## are the variances of ##A## and ##B## (because we assumed ##\langle A \rangle=\langle B \rangle=0##), i.e.,
$$P(\lambda)=\lambda^2 \Delta B^2 + \lambda \langle \mathrm{i} [\hat{A},\hat{B}]\rangle + \Delta A^2 \geq 0$$
für ##\lambda \geq 0##. Now ##P(\lambda)## is a real quadratic polynomial and thus its discriminant fulfills,
$$\frac{1}{4} \langle \mathrm{i} [\hat{A},\hat{B}] \rangle^2 - \Delta A^2 \Delta B^2 \leq 0,$$
and from this finally
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|,$$
which is the Heisenberg-Robertson uncertainty relation.

 

[Jamister]

You can prove it for any state. Let ##\hat{A}## and ##\hat{B}## be the self-adjoint operators representing observables and assume for simplicity that ##\langle A \rangle=\langle B \rangle=0##, where ##\langle \hat{A} \rangle=\mathrm{Tr}(\hat \rho \hat{A})##. Since ##\hat{\rho}## is a positive semi-definite self-adjoint operator there's a complete orthonormalized eigenbasis ##|u_n \rangle##, i.e.,
$$\hat{\rho}=\sum_n p_n |u_n \rangle \langle u_n |.$$
Thus any expectation value reads
$$\langle A \rangle = \mathrm{Tr} (\hat{\rho} \hat{A}) = \sum_{n} \langle u_n| \rho_n \hat{A} |u_n \rangle = \sum_n P_n \langle u_n |\hat{A}| u_n \rangle.$$
Now apply this to the self-adjoint operator
$$\hat{C}=(\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B}), \quad \lambda \in \mathbb{R}.$$
First of all
$$\langle u_n|\hat{C}| u_n \rangle=\langle (\hat{A}+\mathrm{i} \lambda \hat{B}) u_n|(\hat{A}+\mathrm{i} \lambda \hat{B}) u_n \rangle \geq 0.$$
And since ##P_n \geq 0## for all ##n## you also have
$$P(\lambda)=\langle (\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B}) \rangle \geq 0.$$
Multiplying this out gives
$$P(\lambda) = \langle A^2 \rangle + \langle \mathrm{i} [\hat{A},\hat{B}] \rangle \lambda + \lambda^2 \langle \hat{B}^2 \rangle.$$
Now ##\langle A^2 \rangle=\Delta A^2## and ##\langle B^2 \rangle=\Delta B^2## are the variances of ##A## and ##B## (because we assumed ##\langle A \rangle=\langle B \rangle=0##), i.e.,
$$P(\lambda)=\lambda^2 \Delta B^2 + \lambda \langle \mathrm{i} [\hat{A},\hat{B}]\rangle + \Delta A^2 \geq 0$$
für ##\lambda \geq 0##. Now ##P(\lambda)## is a real quadratic polynomial and thus its discriminant fulfills,
$$\frac{1}{4} \langle \mathrm{i} [\hat{A},\hat{B}] \rangle^2 - \Delta A^2 \Delta B^2 \leq 0,$$
and from this finally
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|,$$
which is the Heisenberg-Robertson uncertainty relation.

but you proved it from the beginning. how do you prove it by using the assumption from pure states, and not repeating the all process again

 

[Jamister]

I wouldn't expect the uncertainty in any observable to get smaller by adding statistical uncertainty to the already existing quantum uncertainty in the position and momentum. It can only increase.

There's something said about this on page 16 here: https://www.univie.ac.at/nuhag-php/dateien/talks/1073_NuHAG_WPI_2008.pdf

I don't think it's obvious

 

[PeroK]

but you proved it from the beginning. how do you prove it by using the assumption from pure states, and not repeating the all process again

You only get one proof for free! If you want another one, you have to show some effort yourself.

 

[vanhees71]

but you proved it from the beginning. how do you prove it by using the assumption from pure states, and not repeating the all process again

For pure states the proof is almost the same. You just set one of the ##p_n=1## and all others to 0.