# How reduced density matrix obtained from the matrix.

• I
• munirah
In summary, the reduced density matrix of a bipartite system can be obtained by tracing over the indices of the other subsystem in the composite density matrix, where the indices correspond to the basis of each subsystem. This results in a reduced matrix for each subsystem, with the same number of rows and columns as the original matrix.

#### munirah

Can any expert help me in explaining how this example below get the reduced density matrix from the density matrix in bipartite system.
$$\rho =\frac{1}{4}\begin{pmatrix} 1 & 1 & cos(\frac{\alpha}{2})-sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{2})+sin(\frac{\alpha}{2}) \\ 1 & 1 & cos(\frac{\alpha}{2})-sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{2})+sin(\frac{\alpha}{2}) \\ cos(\frac{\alpha}{2})-sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{2})-sin(\frac{\alpha}{2}) & 1-sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{4}) \\ cos(\frac{\alpha}{2})+sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{2})+sin(\frac{\alpha}{2}) & cos(\frac{\alpha}{4}) & 1+sin(\alpha)\end{pmatrix}$$

For the subsystems, this yields, as below.

$$\rho_\text{A}=Tr_\text{B}(\rho)=\frac{1}{2}\begin{pmatrix}1 & cos(\frac{\alpha}{2}) \\ cos(\frac{\alpha}{2}) & 1\end{pmatrix}$$

and

$$\rho_\text{B}=Tr_\text{A}(\rho)=\frac{1}{2}\begin{pmatrix}1-\frac{1}{2}sin(\alpha) & cos^2(\frac{\alpha}{2}) \\ cos^2(\frac{\alpha}{2}) & 1+\frac{1}{2}sin(\alpha)\end{pmatrix}$$

How the reduced to each system A and B obtained?

If you have a composite density of the form

$\rho_{i j a b}$

where $i, j$ refer to a basis for system A and $a, b$ refer to a basis for system B, then

$(\rho_A)_{i j} = \sum_a \rho_{i j a a}$

$(\rho_B)_{a b} = \sum_i \rho_{i i a b}$

Your 4x4 matrix doesn't clearly indicate which indices are for $A$ and which are for $B$, but I assume you're doing something like:
1. The element with row=1, column=1 corresponds to $i=1, j=1, a=1, b=1$
2. The element with row=1, column=2 corresponds to $i=1, j=1, a=2, b=1$
3. The element with row=2, column=1 corresponds to $i=1, j=1, a=1, b=2$
4. The element with row=2, column=2 corresponds to $i=1, j=1, a=2, b=2$
5. The element with row=1, column=3 corresponds to $i=2, j=1, a=1, b=1$
6. The element with row=1, column=4 corresponds to $i=2, j=1, a=2, b=1$
7. The element with row=2, column=3 corresponds to $i=2, j=1, a=1, b=2$
8. The element with row=2, column=4 corresponds to $i=2, j=1, a=2, b=2$
9. etc.
(or something like that)

munirah
stevendaryl said:
If you have a composite density of the form

$\rho_{i j a b}$

where $i, j$ refer to a basis for system A and $a, b$ refer to a basis for system B, then

$(\rho_A)_{i j} = \sum_a \rho_{i j a a}$

$(\rho_B)_{a b} = \sum_i \rho_{i i a b}$

Your 4x4 matrix doesn't clearly indicate which indices are for $A$ and which are for $B$, but I assume you're doing something like:
1. The element with row=1, column=1 corresponds to $i=1, j=1, a=1, b=1$
2. The element with row=1, column=2 corresponds to $i=1, j=1, a=2, b=1$
3. The element with row=2, column=1 corresponds to $i=1, j=1, a=1, b=2$
4. The element with row=2, column=2 corresponds to $i=1, j=1, a=2, b=2$
5. The element with row=1, column=3 corresponds to $i=2, j=1, a=1, b=1$
6. The element with row=1, column=4 corresponds to $i=2, j=1, a=2, b=1$
7. The element with row=2, column=3 corresponds to $i=2, j=1, a=1, b=2$
8. The element with row=2, column=4 corresponds to $i=2, j=1, a=2, b=2$
9. etc.
(or something like that)
thankyou for helping me. really appreciate.
Sorry, I want to ask more either the notation for
$(\rho_A)_{i j} = \sum_a \rho_{i j a a}$

or

$(\rho_A)_{i j} = \sum_a \rho_{i j a b}$ ?

Thank you very much

munirah said:
thankyou for helping me. really appreciate.
Sorry, I want to ask more either the notation for
$(\rho_A)_{i j} = \sum_a \rho_{i j a a}$

or

$(\rho_A)_{i j} = \sum_a \rho_{i j a b}$ ?

Thank you very much

The first one is correct. That's sort of obvious, because the second one has a free index, $b$ on the right-hand side, but not on the left-hand-side.

munirah
stevendaryl said:
The first one is correct. That's sort of obvious, because the second one has a free index, $b$ on the right-hand side, but not on the left-hand-side.

stevendaryl, can you give me the example according what you explain. I already try to understand. please

You have to give the context. There's no way to guess from just throwing a matrix in the discussion what it means! There must be a clear definition in the textbook/problem sheet given. Otherwise it's just useless!

munirah said:
stevendaryl, can you give me the example according what you explain. I already try to understand. please

I don't understand what it is that you don't understand. The density matrix for a two-component system has 4 indices: two for the first component, and two for the second component. To form the reduced matrix for one component, you "trace" over the indices for the other component. A trace means setting the two indices to the same value and summing over all possibilities.

Your original matrix doesn't clearly identify those 4 indices, so it's not possible to form the reduced matrix without additional information. That was probably present, if you saw the problem in a textbook.

From your 16-element matrix, what I'm assuming the components of $\rho_{ijab}$ are:
1. $\rho_{1111} = \frac{1}{4}$
2. $\rho_{1112} = \frac{1}{4}$
3. $\rho_{1121} = \frac{1}{4}$
4. $\rho_{1122} = \frac{1}{4}$
5. $\rho_{1211} = \frac{1}{4}(cos(\frac{\alpha}{2}) - sin(\frac{\alpha}{2}))$
6. $\rho_{1212} = \frac{1}{4}(cos(\frac{\alpha}{2}) + sin(\frac{\alpha}{2}))$
7. $\rho_{1221} = \frac{1}{4}(cos(\frac{\alpha}{2}) - sin(\frac{\alpha}{2}))$
8. $\rho_{1222} = \frac{1}{4}(cos(\frac{\alpha}{2}) + sin(\frac{\alpha}{2}))$
Etc.

So
$(\rho_A)_{ij} = \rho_{ij11} + \rho_{ij22}$
$(\rho_B)_{ab} = \rho_{11ab} + \rho_{22ab}$

munirah and vanhees71
tqvm stevendaryl. I understand now