Density Matrix of Multiple Qubits

[thatboi]

Hey all,
I am having trouble relating probabilities with the density matrix of multiple qubits. Consider we have a system of 3 qubits: the first qubit is in the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$ and the remaining 2 qubits are prepared in the state described by the density matrix $\rho = a_{1}\ket{\alpha_{1}}\bra{\alpha_{1}}\otimes\ket{\beta_{1}}\bra{\beta_{1}} + a_{2}\ket{\alpha_{2}}\bra{\alpha_{2}}\otimes\ket{\beta_{2}}\bra{\beta_{2}}$
where we only know that $\braket{\alpha_{i}|\alpha_{i}} = \braket{\beta_{i}|\beta_{i}} = 1$ for $i = 1,2$ and $a_{1} + a_{2} = 1$.
Now suppose we form the density matrix $\rho_{tot}$ of all 3 qubits.

My question is: From $\rho_{tot}$, how do I obtain the probability of the first qubit to be in state $\ket{0}$, which we know is $\frac{1}{2}$. Normally I would think of performing a partial trace but I'm not sure of what to take the partial trace over in this case since we do not have further information on the other 2 qubits.

 

[PeterDonis]

From $\rho_{tot}$, how do I obtain the probability of the first qubit to be in state $\ket{0}$, which we know is $\frac{1}{2}$.

You don't have to obtain that probability from $\rho_{tot}$ because you already know it: you specified it when you specified the state of the first qubit. Since the state you specified is a product state, i.e., the first qubit is not entangled with the other two, the state of the other two qubits doesn't matter.

 

[thatboi]

You don't have to obtain that probability from $\rho_{tot}$ because you already know it: you specified it when you specified the state of the first qubit. Since the state you specified is a product state, i.e., the first qubit is not entangled with the other two, the state of the other two qubits doesn't matter.

Hi,
I agree that we don't need $\rho_{tot}$, but we should technically be able to obtain that probability from $\rho_{tot}$ still right. I was just wondering how I could get that value from $\rho_{tot}$, such as from the diagonal values?

 

[PeterDonis]

we should technically be able to obtain that probability from $\rho_{tot}$ still right.

You obtain it the way I just said: by recognizing that the total state is a product state, and therefore you don't need to know anything about the states of the other qubits. The "partial trace" operation for this case just means ignoring the other qubits and using the known state of the first qubit.

I was just wondering how I could get that value from $\rho_{tot}$, such as from the diagonal values?

You can of course view the "partial trace" operation, described above, as reading off the part of $\rho_{tot}$ that refers to the first qubit and using that and ignoring the rest. The answer is the same.

 

[thatboi]

You obtain it the way I just said: by recognizing that the total state is a product state, and therefore you don't need to know anything about the states of the other qubits. The "partial trace" operation for this case just means ignoring the other qubits and using the known state of the first qubit.You can of course view the "partial trace" operation, described above, as reading off the part of $\rho_{tot}$ that refers to the first qubit and using that and ignoring the rest. The answer is the same.

Perhaps I misstated my problem, but my confusion originally arose from this post: https://quantumcomputing.stackexcha...ap-test-and-density-matrix-distinguishability
Specifically by the statement in the accepted answer: "We're interested by the diagonal coefficients of $\rho_{3}$ that can be written as $\ket{0,i,j}\bra{0,i,j}$. Summing them would give us the probability of measuring |0⟩." I'm not sure what assumptions were made about $\ket{i,j}$, but I wanted to know the reasoning behind this statement.

 

[PeterDonis]

my confusion originally arose from this post

Which is not about the kind of state you described in the OP. It's about multi-qubit states that are not product states, with various quantum computing gates applied to them. Which kind of state do you really want to ask about?

 

[thatboi]

Which is not about the kind of state you described in the OP. It's about multi-qubit states that are not product states, with various quantum computing gates applied to them. Which kind of state do you really want to ask about?

I would like a clarification on my question regarding the SWAP gate that I mentioned in my previous reply.

 

[PeterDonis]

"We're interested by the diagonal coefficients of $\rho_{3}$ that can be written as $\ket{0,i,j}\bra{0,i,j}$. Summing them would give us the probability of measuring |0⟩." I'm not sure what assumptions were made about $\ket{i,j}$, but I wanted to know the reasoning behind this statement.

It's an obvious consequence of the definition of a density matrix. The coefficients described are by definition the ones whose sum gives the probability of measuring $\ket{0}$.

 

[Demystifier]

Normally I would think of performing a partial trace but I'm not sure of what to take the partial trace over in this case since we do not have further information on the other 2 qubits.

You know that the other two qubits (which I will call "the rest") have a density matrix, and trace of any density matrix is 1. That's all what you need. More explicitly, your full system has a density matrix of the form
$$\rho=\rho_{\rm first}\otimes\rho_{\rm rest}$$
where $\rho_{\rm first}=|\psi\rangle\langle\psi|$ and $\rho_{\rm rest}$ describes the other two qubits. The reduced density matrix of the first subsystem is therefore
$$\rho_{\rm first \; (reduced)}={\rm Tr}_{\rm rest}\,\rho
= \rho_{\rm first} ({\rm Tr}_{\rm rest}\rho_{\rm rest}) = \rho_{\rm first}$$
The result is trivial because the full state is a product, i.e. there is no entanglement between the first subsystem and the rest.

 

[Structure seeker]

The total density matrix will be $0.5 \begin{pmatrix} \rho & \rho \\ \rho & \rho \end{pmatrix}$