# Reduced density matrix does not reproduce expectation values

It seems I have found a counter example of the theorem that states that the reduced density matrix reproduces the expectation values in that corresponding subsystem:

We set $\bar{h}=c=1$. The system is a $1+1$ dimensional periodic lattice with only two sites. With Hamiltonian $$H=\frac{1}{\varepsilon}\sum_{j=0}^{N-1}\left(m\varepsilon(\psi_{-,j}^\dagger\psi_{+,j}+\psi_{+,j}^\dagger\psi_{-,j})\right.\\ \left.-\frac{1}{2}i\left(\psi_{+,j}^\dagger\psi_{+,j+1}-\psi_{+,j+1}^\dagger\psi_{+,j}-(\psi_{-,j}^\dagger\psi_{-,j+1}-\psi_{-,j+1}^\dagger\psi_{-,j})\right)\right).$$ and boundary conditions

$\psi_{+,N}=e^{2\pi i \theta}\psi_{+,0}$, and $\psi_{-,N}=e^{2\pi i \theta} \psi_{-,0}$.

Notice that the $\psi$ field is dimensionless. The wavefunction obeys the Dirac equation, giving the dispersion relation $$\omega_k^2=m^2+\tilde{k}(k)^2=m^2+\tilde{k}(k)^2,$$ with $m$ the mass, $\tilde{k}(k)=\frac{1}{\varepsilon}\sin(\pi k)$ and $\varepsilon$ the lattice spacing.

The operators $\psi_{\pm,j}$ satisfy the fermionic anticommutation relations, so they correspond to annihilation operators. So we have four annihilation operators. Then we set for notational convenience $k:=\tilde{k}(\theta)=-\tilde{k}(\theta+1)$ and $\omega :=\omega_{\theta}=\omega_{\theta+1}$. We denote a ket state in the following convention: $$|a,b,c,d\rangle=(\psi^\dagger_{-,1})^d(\psi^\dagger_{-,0})^c(\psi^\dagger_{+,1})^b(\psi^\dagger_{+,0})^a|0,0,0,0\rangle,$$ with $a,b,c,d\in{0,1}$. Consider the state $$\nonumber |0\rangle:=\frac{1}{2 \omega}\left(m|0,0,1,1\rangle+k e^{i\pi\theta}|0,1,0,1\rangle+\omega|0,1,1,0\rangle\\-\omega|1,0,0,1\rangle-k e^{-i\pi\theta}|1,0,1,0\rangle+m|1,1,0,0\rangle\right),$$ which is actually the groundstate of the Hamiltonian (but that is not relevant). The corresponding density matrix is $\rho=|0\rangle\langle0|$. At this groundstate we compute the correlation matrix $C=\langle\eta^\dagger\eta\rangle$ with $\eta=(\psi_{+,0},\psi_{+,1},\psi_{-,0},\psi_{-,1})$. This is a straightfoward calculation but care needs to be taken when permuting the operators since they anticommute. The result is: $$C=\frac{1}{2\omega}\begin{pmatrix} \omega & -k e^{i\pi\theta} & -m &0 \\ -k e^{-i\pi\theta} & \omega &0 & -m \\ -m &0 &\omega&k e^{i\pi\theta}\\0&-m&k e^{-i\pi\theta}&\omega\end{pmatrix}$$ At this point we can consider the part of this correlation matrix corresponding to the operators working on the $0^{th}$ site: $$C_0=\frac{1}{2\omega}\begin{pmatrix} \omega & -m \\ -m &\omega\end{pmatrix}.$$ These expectation values should be reproduced by the reduced density matrix of this subsystem: $$\rho_0=\frac{1}{4\omega^2}\left((m^2+\omega^2)(|0,1\rangle\langle0,1|+|1,0\rangle\langle1,0|)+k^2(|1,1\rangle\langle1,1|+|0,0\rangle\langle0,0|)\right),$$ obtained by tracing out the second and fourth quantum number in the total density matrix. We see that the diagonal components of $C_0$ are indeed reproduced by the reduced density matrix, using $\omega^2=m^2+k^2$. However $Tr(\rho_0\psi_{+,0}^\dagger\psi_{-,0})=0\neq\frac{-m}{2\omega}$.

So somewhere I make a mistake, either in reducing the density matrix or in reducing $C$ to $C_0$. But I don't see where the fault is.

• 1) You have several typos, e.g. your definition of the ground state ket. 2) Your correlation matrix $C$ is clearly wrong. The diagonal elements of this matrix should describe probabilities (i.e. the probability that a given mode is occupied). Yet the elements of the matrix you have written are probability amplitudes, e.g. the parameters $m$, $\omega$ etc. appearing in the state. As you know, probabilities are given by amplitudes squared, like in your (presumably correct) expression for $\rho_0$. – Mark Mitchison Feb 3 '17 at 13:47
• @MarkMitchison 1) Thanks for your comment, I believe the typo's are gone now. 2) I don't see why $C$ is wrong. The diagonal entries are precisely the expectation vanlues of the counting operators. So the top left entry is the expectation value of the first quantum number: only the last 3 kets in the groundstate have a $\psi_{+,0}$ "particle", so this expectation value is just the sum of the square of the modulus of the three coefficients: $\frac{1}{4\omega^2}(\omega^2+k^2+m^2)=\frac{1}{2}$ using $\omega^2=m^2+k^2$. – DionH Feb 4 '17 at 10:05
• It's just simple power counting. The state $|0\rangle$ is linear in the amplitudes $m/\omega, k/\omega$ etc. The correlation matrix $C$ contains "two factors" of $|0\rangle$, since it's matrix of expectation values. Therefore, $C$ is quadratic in the amplitudes. There's just no way that something like $m/\omega$ could appear. If you go through the calculation again carefully you will see. – Mark Mitchison Feb 4 '17 at 14:05
• @MarkMitchison The third entry in the first row of $C$ is $\langle\psi^\dagger_{+,0}\psi_{-,0}\rangle$, which, effectively, is the expectation value of the operator that annihilates the third quantum number, and then creates the first quantum number. This operation will give a non zero outcome only on the first and third ket. Keeping in mind the anti commutation relations, we get $\psi^\dagger_{+,0}\psi_{-,0}|0\rangle=\frac{1}{2\omega}(-\omega|1,1,0,0\rangle+m|1,0,1,0\rangle)$. Hence if we multiply this with the groundstate bra, we get $\frac{1}{4\omega^2}(-2m\omega)=\frac{-m}{2\omega}$. – DionH Feb 4 '17 at 14:32
• Got it. You've convinced me that your C is ok. Then I think the problem is with your reduced state. Indeed, you are missing terms like $Tr_1(|0011\rangle\langle 0110|)=|01\rangle\langle 10|$. These will recover the non-zero off-diagonal elements of the correlation matrix. – Mark Mitchison Feb 4 '17 at 15:01

It turns out I missed an anticommutator step in my calculation: $$\begin{split} Tr_1(|0110\rangle\langle1100|)&=(id×⟨00|)ψ_{+,1}ψ^†_{−,0}ψ^†_{+,1}|0000⟩⟨0000|ψ_{+,0}ψ_{+,1}ψ^†_{+,1}(id×|00⟩)\\&=−|01⟩⟨10| \end{split}$$ because I need to permute the two most left operators. Hence the reduced density matrix contains two extra terms $\frac{-m}{2\omega}|01⟩⟨10|$ and $\frac{-m}{2\omega}|10⟩⟨01|$, yielding the desired off-diagonal entries. This trouble could have been avoided if I took my ket state convention such that the site 1 operators are all on the left.