# How Is the Mixed State Density Matrix Represented in Quantum Computing?

• I
• thatboi
thatboi
Hi all,
I am having trouble visualizing the matrix representation of the mixed density matrix from the following post (specifically from the accepted answer): https://quantumcomputing.stackexcha...ap-test-and-density-matrix-distinguishability
That is, for ##\rho_3=\frac{1}{4}\frac{1}{2^{2n}}\sum_{i,j}\left(\sum_{a,b}|a,i,j\rangle\langle b,i,j|+\sum_{a,b}(-1)^b|a,i,j\rangle\langle b,j,i|+\sum_{a,b}(-1)^a|a,j,i\rangle\langle b,i,j|+\sum_{a,b}(-1)^{a\oplus b}|a,j,i\rangle\langle b,j,i|\right)##. The user makes the statement: "We're interested by the diagonal coefficients of ##\rho_{3}## that can be written as |0,i,j⟩⟨0,i,j|. Summing them would give us the probability of measuring |0⟩. " I just wanted to confirm if ##\rho_{3}## was still an 8x8 matrix, and if so, is there an implicit assumption that the ##{i,j}## states form an orthonormal basis of their 4x4 Hilbert space?
Thanks.

##\rho_3## should be the "partial trace" of the original density matrix over the ##i##, ##j## subsystem. It's sort of like how the Ricci tensor isn't the full trace of the Riemann curvature tensor (which would give the scalar curvature.) So ##\rho_3## should be a 2 by 2 matrix, and at a practical level you could think of ##\rho_3## (as you defined it) as the density matrix that completely characterizes observables involving the first spin index (##a##).

Couchyam said:
##\rho_3## should be the "partial trace" of the original density matrix over the ##i##, ##j## subsystem. It's sort of like how the Ricci tensor isn't the full trace of the Riemann curvature tensor (which would give the scalar curvature.) So ##\rho_3## should be a 2 by 2 matrix, and at a practical level you could think of ##\rho_3## (as you defined it) as the density matrix that completely characterizes observables involving the first spin index (##a##).
Hi,
why is ##\rho_{3}## a 2x2 matrix? I thought the expression for ##\rho_{3}## was just a sum over several outer products ##\ket{a,i,j}\bra{a,i,j}## where each term describes a system of 3 spins.

thatboi said:
Hi,
why is ##\rho_{3}## a 2x2 matrix? I thought the expression for ##\rho_{3}## was just a sum over several outer products ##\ket{a,i,j}\bra{a,i,j}## where each term describes a system of 3 spins.
The complete density matrix would have no repeated indices, or would include a sum over both ##V## indices as well as ##V^*## indices:
$$\hat\rho_0 = \rho_{a,i,j; b,k,\ell} |a,i,j\rangle\langle b, k, \ell| \equiv \sum_{a,i,j,b,k,\ell} \rho_{a,i,j; b,k,\ell} |a, i, j\rangle\langle b, k, \ell|$$
where ##\rho_{a,i,j; b,k,\ell} = \rho^*_{b,k,\ell; a,i,j}##.

thatboi said:
why is ##\rho_{3}## a 2x2 matrix?
Because it describes one qubit, i.e., it is a density matrix on a 2-dimensional Hilbert space.

The partial trace operation reduces the size of the original density matrix, which will be a 2n x 2n matrix, where n is the number of qubits.

Couchyam said:
The complete density matrix would have no repeated indices, or would include a sum over both ##V## indices as well as ##V^*## indices:
$$\hat\rho_0 = \rho_{a,i,j; b,k,\ell} |a,i,j\rangle\langle b, k, \ell| \equiv \sum_{a,i,j,b,k,\ell} \rho_{a,i,j; b,k,\ell} |a, i, j\rangle\langle b, k, \ell|$$
where ##\rho_{a,i,j; b,k,\ell} = \rho^*_{b,k,\ell; a,i,j}##.
So how what would the matrix of say, ##\rho_0=\frac{1}{2^{2n}}\sum_{i,j}|0,i,j\rangle\langle0,i,j|## look like? I'm not sure what basis states to pick to represent this as a 2x2 matrix because of the presence of the ##\ket{i,j}## terms.

thatboi said:
So how what would the matrix of say, ##\rho_0=\frac{1}{2^{2n}}\sum_{i,j}|0,i,j\rangle\langle0,i,j|## look like? I'm not sure what basis states to pick to represent this as a 2x2 matrix
It's not a 2x2 matrix because it's not representing just one qubit. It's representing all the qubits. So it's a 2n x 2n matrix.

For the first qubit, you would use the ##\ket{0}##, ##\ket{1}## basis. For the other qubits, it doesn't matter what basis you pick; as I'm reading the stack exchange thread, ##i## and ##j## represent any pair of basis states you like for the other qubits.

PeterDonis said:
It's not a 2x2 matrix because it's not representing just one qubit. It's representing all the qubits. So it's a 2n x 2n matrix.

For the first qubit, you would use the ##\ket{0}##, ##\ket{1}## basis. For the other qubits, it doesn't matter what basis you pick; as I'm reading the stack exchange thread, ##i## and ##j## represent any pair of basis states you like for the other qubits.
Sorry, I think I am misunderstanding something: isn't ##\rho_{0}## an example of a term that would appear in the expression of ##\rho_{3}##? But it was claimed that ##\rho_{3}## was a 2x2 matrix.

thatboi said:
isn't ##\rho_{0}## an example of a term that would appear in the expression of ##\rho_{3}##?
No. ##\rho_0## is the initial state. ##\rho_3##, at least as that term was used in post #2, is the partial trace of the final state over all of the qubits except the first. (In the stack exchange thread, ##\rho_3## is just the final state itself, without the partial trace operation; but the rest of that answer basically describes the partial trace operation, since that's what you do to compute the probability of measuring the first qubit to be in state ##\ket{0}##.)

You appear to be confused by the fact that the expression ##\ket{0, i, j} \bra{0, i, j}## appears in both the formula for ##\rho_0## and the discussion of the partial trace operation. That is true, but it doesn't imply what you stated in the quote above. If we want to be really pedantic about how to compute the probability of measuring the first qubit to be in state ##\ket{0}##, we would say that you take the final state that results from the gate operations described, which is a 2n x 2n density matrix, take its partial trace over all the qubits except the first, which gives a 2 x 2 matrix that was called ##\rho_3## in post #2, and then compute the probability for ##\ket{0}## based on that 2 x 2 matrix, which is a straightforward operation. The stack exchange answer, when it talks about the expression ##\ket{0, i, j} \bra{0, i, j}## at the end, is basically combining the last two steps I just described (taking the partial trace and computing the probability of ##\ket{0}##).

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