Calculating Matrix Representation of 11><11 for 2 Qubits

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Discussion Overview

The discussion revolves around calculating the matrix representation of the outer product |11><11| for two qubits. Participants explore the correct formulation of the matrix and the implications of basis vector ordering in this context.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that forming the matrix representation of |1><1| is straightforward, but they seek help with |11><11|, providing an initial incorrect matrix.
  • Another participant suggests that |1> can be represented as (1 0)t, leading to the conclusion that |11> is (1 0 0 0)t, resulting in a matrix with a 1 in the upper left corner and zeros elsewhere.
  • A clarification is provided by a participant who interprets |11> as (0 1 0 1) and <10 as (0 1 1 0), expressing confusion over the outer product having more than one non-zero entry.
  • Another participant proposes that |11< refers to (0 0 0 1) and <10 refers to (0 1 0 0), questioning if this is correct.
  • A later reply emphasizes that the matrix elements depend on the order of the basis vectors, suggesting a specific ordering and providing the resulting matrix for |11><11| as a projection operator in a 4-dimensional space.
  • A participant acknowledges that their issue stemmed from incorrectly ordering the basis vectors, expressing a lingering concern about the necessity of such ordering in relation to nature.

Areas of Agreement / Disagreement

Participants express differing interpretations of the matrix representation and the ordering of basis vectors, indicating that multiple competing views remain and the discussion is unresolved.

Contextual Notes

Participants highlight the dependence of matrix representation on the chosen order of basis vectors, which remains a point of contention and uncertainty.

Getterdog
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Forming the matrix representation of say 1><1 is no problem but how does one calculate the matrix representation of 11><11 ? Is it
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
Any help? thanks jack
 
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i suppose 1> is (1 0)t then 11 is (1 0 0 0)t and the matrix is 1 in the upper left corner and all the rest 0.
 
Clarification

I take that 11>. Is (0 1 0 1 ) and < 10. Refers to ( 0 1 1 0 ) . The outer product as a matrix has more than 1 non zero entry.,so I'm still stuck. Any clarification on the correct way to do this.? Thanks
 
Unless 11< refers to ( 0 0 0 1) and 10< refers to (0 1 0 0 ) ,01< to (0 0 1 0) Is this it??
 
The matrix elements depend on the order of the basis vectors, so you need to choose a way to order them, e.g. (|00>,|01>,|10>,|11>). The matrix of |11><11| with respect to this ordered basis is
\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} This could be anticipated from the fact that |11><11| is a projection operator for a 1-dimensional subspace of the vector space, which is 4-dimensional.

The matrix of any operator T with respect to this ordered basis is
\begin{pmatrix}\langle 00|T|00\rangle & \langle 00|T|01\rangle &\dots & &\\ \langle 01|T|00\rangle & \langle 01|T|01\rangle & \dots \\ \vdots & \vdots & \ddots \\ \end{pmatrix}
For more information, see the https://www.physicsforums.com/showthread.php?t=694922 about the relationship between linear operators and matrices.
 
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If my previous reply didn't register, thanks,my problem was that I didn't order the basis vectors correctly. Now I can proceed , although something bugs me about ordering of the basis, nature doesn't seem to care about humans need to order anything. Thanks
 

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