How reduced density matrix obtained from the matrix.

  • #1
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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?
 

Answers and Replies

  • #2
stevendaryl
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If you have a composite density of the form

[itex]\rho_{i j a b}[/itex]

where [itex]i, j[/itex] refer to a basis for system A and [itex]a, b[/itex] refer to a basis for system B, then

[itex](\rho_A)_{i j} = \sum_a \rho_{i j a a}[/itex]

[itex](\rho_B)_{a b} = \sum_i \rho_{i i a b}[/itex]

Your 4x4 matrix doesn't clearly indicate which indices are for [itex]A[/itex] and which are for [itex]B[/itex], but I assume you're doing something like:
  1. The element with row=1, column=1 corresponds to [itex]i=1, j=1, a=1, b=1[/itex]
  2. The element with row=1, column=2 corresponds to [itex]i=1, j=1, a=2, b=1[/itex]
  3. The element with row=2, column=1 corresponds to [itex]i=1, j=1, a=1, b=2[/itex]
  4. The element with row=2, column=2 corresponds to [itex]i=1, j=1, a=2, b=2[/itex]
  5. The element with row=1, column=3 corresponds to [itex]i=2, j=1, a=1, b=1[/itex]
  6. The element with row=1, column=4 corresponds to [itex]i=2, j=1, a=2, b=1[/itex]
  7. The element with row=2, column=3 corresponds to [itex]i=2, j=1, a=1, b=2[/itex]
  8. The element with row=2, column=4 corresponds to [itex]i=2, j=1, a=2, b=2[/itex]
  9. etc.
(or something like that)
 
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  • #3
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If you have a composite density of the form

[itex]\rho_{i j a b}[/itex]

where [itex]i, j[/itex] refer to a basis for system A and [itex]a, b[/itex] refer to a basis for system B, then

[itex](\rho_A)_{i j} = \sum_a \rho_{i j a a}[/itex]

[itex](\rho_B)_{a b} = \sum_i \rho_{i i a b}[/itex]

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

or

[itex](\rho_A)_{i j} = \sum_a \rho_{i j a b}[/itex] ?

Thank you very much
 
  • #4
stevendaryl
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thankyou for helping me. really appreciate.
Sorry, I want to ask more either the notation for
[itex](\rho_A)_{i j} = \sum_a \rho_{i j a a}[/itex]

or

[itex](\rho_A)_{i j} = \sum_a \rho_{i j a b}[/itex] ?

Thank you very much
The first one is correct. That's sort of obvious, because the second one has a free index, [itex]b[/itex] on the right-hand side, but not on the left-hand-side.
 
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  • #5
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The first one is correct. That's sort of obvious, because the second one has a free index, [itex]b[/itex] on the right-hand side, but not on the left-hand-side.
tqvm stevendaryl for your respon
 
  • #6
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stevendaryl, can you give me the example according what you explain. I already try to understand. please
 
  • #7
vanhees71
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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
stevendaryl
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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 text book.

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

So
[itex](\rho_A)_{ij} = \rho_{ij11} + \rho_{ij22}[/itex]
[itex](\rho_B)_{ab} = \rho_{11ab} + \rho_{22ab}[/itex]
 
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  • #9
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tqvm stevendaryl. I understand now
 

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