# I How reduced density matrix obtained from the matrix.

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1. Jun 19, 2016

### 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?

2. Jun 19, 2016

### stevendaryl

Staff Emeritus
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)

3. Jun 19, 2016

### munirah

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

4. Jun 19, 2016

### stevendaryl

Staff Emeritus
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.

5. Jun 19, 2016

### munirah

6. Jun 20, 2016

### munirah

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

7. Jun 20, 2016

### vanhees71

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!

8. Jun 20, 2016

### stevendaryl

Staff Emeritus
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 text book.

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

9. Jun 21, 2016

### munirah

tqvm stevendaryl. I understand now