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I am looking for a computationally efficient way to minimize the following function. Let $$\Phi(\rho, U) = \text{Tr}_2(U\rho U^\dagger)$$ be a reduced density matrix where $\rho = \overline{\rho}_1 \otimes \rho_2$ is a product density matrix over $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2 \cong \mathbb{C}^2 \otimes \mathbb{C}^2$ such that $\overline{\rho}_1$ is pure and $U \in SU(4)$. The overline is purely to emphasize that $\overline{\rho}_1$ is pure.

Though low dimensional, this reduced density matrix is generally complicated. In particular, given a fixed $\rho$, I am interested in the minimization problem $$1 - \text{Tr}(\Phi(\rho, U)^2). \tag{1}$$ The absolute minimum value of $(1)$ is just the value $0$. Hence, exactly minimizing $(1)$ would be solving for the (assuming nonzero) set of all $U \in SU(4)$ such that $$1 - \text{Tr}(\Phi(U)^2) = 0 \tag{2}$$ where I have dropped the explicit dependence on $\rho$ to emphasize that $\rho$ is fixed. There are many equivalent criterions to $1 - \text{Tr}(\Phi(\rho, U)^2) = 0$, some of which are special results to $2 \times 2$ matrices:

  1. $\Phi(\rho, U)$ is rank $1$,
  2. $\det \Phi(\rho, U) = 0$,
  3. $\text{spec}(\Phi(\rho, U)) = \{0, 1\}$,
  4. $\max \text{spec}(\Phi(\rho, U)) = 1$
  5. $\min \text{spec}(\Phi(\rho, U)) = 0$,
  6. ...

I have been trying to wrap my head around the best way to solving $(2)$.

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    $\begingroup$ Why do you care about this quantity? $\endgroup$ Commented Mar 15 at 4:58
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    $\begingroup$ Like @NorbertSchuch mentioned, please add motivation behind what you are trying to do or need to do. This will help people answer your question better. $\endgroup$
    – FDGod
    Commented Mar 15 at 5:41
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    $\begingroup$ What is $\rho_1(U)$? How is the dependence of $\rho$ on this $U$ defined? And why has $\rho$ a subscript $_1$ ? $\endgroup$ Commented Mar 15 at 7:09
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    $\begingroup$ Sorry, I've made the post more precise now. @JosBergervoet $\endgroup$ Commented Mar 15 at 7:29
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    $\begingroup$ OK, sorry for being inquisitive, but if your space is $\mathbb{C}^2 \otimes \mathbb{C}^2$, then you can just say $U$ is in SU(4), but you say $U \in SU(\text{dim}\mathcal{H})$, do you want it for arbitrary dimensions? $\endgroup$ Commented Mar 15 at 7:51

1 Answer 1

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In your setup, the choice $U={\bf 1}$ of course suffices to minimize your function $$1 - \text{Tr}(\Phi(\rho, U)^2), \tag{1}$$ because we then have $\Phi(\rho, U) = {\rm Tr}_2(\rho) = \overline{\rho}_1,$ which gives you the minimum value of $0$ for function $(1)$ because $\overline{\rho}_1$ is pure. To find all solutions it might be better to first bring $\rho$ to standard form: $$ \rho_{\rm s} = U_{\rm s}\ \rho \ U_{\rm s}^\dagger = \begin{pmatrix} \lambda_1 \\ & \lambda_2 \\ & & 0 \\ & & & 0 \end{pmatrix}. $$ After that it is easy to see subsequent transformations that are allowed:

$$ \begin{align} U = &\ U_C\ U_B\ U_A = \begin{pmatrix} U_2 & 0 \\ 0 & U_2 \end{pmatrix} \Big( U_1 \otimes {\bf 1} \Big) \begin{pmatrix} {\bf 1} & 0 \\ 0 & U_0 \end{pmatrix} \\[5pt] = &\ \begin{pmatrix} U_2 & 0 \\ 0 & U_2 \end{pmatrix} \ \underbrace{\begin{pmatrix} \alpha \ {\bf 1} & \beta \ {\bf 1} \\ -\beta^* \ {\bf 1} & \alpha^* \ {\bf 1} \end{pmatrix}}_{|\alpha|^2+|\beta|^2=1} \ \begin{pmatrix} {\bf 1} & 0 \\ 0 & U_0 \end{pmatrix}, \end{align} $$ where $U_0, U_1$ and $U_2$ are $SU(2)$ matrices. They keep $(1)$ invariant because $U_0$ clearly acts only on the null-space of $\rho_{\rm s}$, the $U_2$ on the left will become invisible if ${\rm Tr}_2$ is taken, and subsequently the $U_1$, which works onky in space $\mathcal{H}_1$, will be invisible when the purity is computed for the remaining state in $\mathcal{H}_1$ by the trace of the square. So we conclude that the allowed transformations for $\rho$ are at least $U_C\ U_B\ U_A \ U_{\rm s}$, where the first three factors give us 9 real parameters. But is that all?

minimum

As mentioned in the comments, one might expect heuristically a 14-parameter family of transformations to be allowed, since full SU(4) is a 15 real-parameter group and we have only 1 real-valued function as a constraint. This would hold if the constraint has a finite first derivative, but not if $(1)$ is already in its minimum. In general, the parameter set where an extremum occurs can have lower dimension than the set for other constant values of the constraint (see drawing, where the minimum is one point whereas for other values we have a circle).

To prove that in the case of function $(1)$ the number of parameters for the minimum at $0$ and for other (nonzero) values are 9 and 14 respectively, we can parametrize $U$ in the neighborhood of unity: $$ \begin{align} U &= \text{exp}(z\ T) = {\bf 1}+z\ T+\frac{z^2}2 T^2 + O(z^3), \\ U^\dagger = U^{-1} &= \text{exp}(-z\ T) = {\bf 1}-z\ T+\frac{z^2}2 T^2 + O(z^3), \end{align}$$ where $T$ is an arbitrary element of the Lie algebra, parametrized by 15 real parameters, $a,b,c,d,e,f,g,h, q ,j,k,l,m,n, p,$ as: $$ \left( \begin{array}{cccc} i q+i p & j+i k & a+i b & c+i d \\ -j+i k & -i q + i p & e+i f & g+i h \\ -a+i b & -e+i f & i l-i p & m+i n \\ -c+i d & -g+i h & -m+i n & -i l-i p \\ \end{array} \right) $$ and we have to do all calculation up to 2nd order in $z$. To continue by hand, it might help to use the factorization given above, but by using a math engine we can also work with the entire matrix at once. See details below, where we find the result for the final purity in space $\mathcal{H}_1$ as: $$ {\rm \gamma_1} = {\rm Tr}\big({\rm Tr}_2(U\ \rho_{\rm s} \ U^{-1})^2\big) = (\lambda_1+\lambda_2)^2 -2 z^2\ C_2, \quad\text{with:} $$ $$ C_2 = (c^2 + d^2)\lambda_1^2 + \big(c^2 + d^2 + e^2 + f^2 + (a - g)^2 + (b - h)^2\big)\lambda_1\lambda_2 + (e^2 + f^2)\lambda_2^2 $$ Our function $(1)$ is just $1-\gamma_1$ and to have the 2nd order term vanish we need $c, d, e, f, a-g,$ and $b-h$ all to be $0$, so we lose 6 of the 15 degrees of freedom of $SU(4)$, showing that indeed only a 9-parameter family of transformations is allowed to keep $(1)$ minimized!

If, on the other hand, we want $(1)$ to have a fixed value slightly higher than $0$, then we only need to fullfil one relation, e,g, by choosing: $$ a = \sqrt{-(b-h)^2-\frac{\lambda_1 \left(c^2+d^2\right)}{\lambda_2}-c^2-d^2-\frac{\lambda_2 \left(e^2+f^2\right)}{\lambda_1}-e^2-f^2+\frac{X}{\lambda_1 \lambda_2}} $$ which immediately gives $C_2=X$, so we are then left with a 14-parameter family of $SU(4)$ transformations which for a (fixed) small value of $z$ will all keep $(1)$ constant at the value $2z^2 X$.

Calculation details:

Clear["Global`*"]

  (*  Tr2 definition, and some assumptions, mainly to declare some parameters real!  *)
Tr2[x_]={{x[[1,1]]+x[[2,2]], x[[1,3]]+x[[2,4]]},{x[[3,1]]+x[[4,2]],x[[3,3]]+x[[4,4]]}}
As=Assumptions->{a<1,b<1,c<1,d<1,e<1,f<1,g<1,h<1,j<1,k<1,l<1,m<1,n<1,p<1, q<1, l1<1,l2<1} 
  
  (* now construct some matrices:  *)
rho = {{l1,0,0,0},{0, l2,0,0},{0,0,0,0},{0,0,0,0}}    (*   <-- standard form: rho_s  *)
id4= {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}
Tc ={{0,0,a+b I, c+d I},{0,0,e+f I,g +h I},{0,0,0,0},{0,0,0,0}}
Tt =Simplify[-Conjugate[Transpose[Tc]], As]+Tc         (*  off-diag. 2x2 blocks  *)
TU2={{q I,j+I k,0,0},{-j+I k,-q I,0,0},{0,0,0,0},{0,0,0,0}}
TU0={{0,0,0,0},{0,0,0,0},{0,0,l I,m+I n},{0,0,-m+I n,-l I}}
Tp ={{p I,0,0,0},{0,p I,0,0},{0,0,-p I,0},{0,0,0,-p I}}
T = Tt+TU0+TU2+Tp;  MatrixForm[T]     (*  <-- check if it worked?  *)

This does indeed give us the desired matrix $T$ from the Lie algebra :

Out[13]//MatrixForm = I p + I q    j + I k      a + I b      c + I d

                      -j + I k     I p - I q    e + I f      g + I h

                      -a + I b     -e + I f     I l - I p    m + I n

                      -c + I d     -g + I h     -m + I n     -I l - I p

Now we do the transformation and the trace calculations, all up to 2nd order in $z$ correct:

U = id4+z T + z^2/2 T.T                    (* compute U up to 2nd order in z,  *)
U = FullSimplify[U, As];  Ui= (U/.z->-z)   (* and its inverse Ui. Now compute:  *)

UrhoUi = FullSimplify[Normal[Series[ U.rho.Ui, {z,0,2} ]] ,As]
Tr2T = FullSimplify[Normal[Series[ Tr2[UrhoUi], {z,0,2} ]] ,As]
Pur1 = FullSimplify[Normal[Series[ Tr[Tr2T.Tr2T], {z,0,2}]], As]
                  2         2    2    2
Out[18]= (l1 + l2)  + (-2 (c  + d ) l1  -

            2    2    2    2          2          2              2    2    2   2
>       2 (c  + d  + e  + f  + (a - g)  + (b - h) ) l1 l2 - 2 (e  + f ) l2 ) z
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