New answers tagged density-matrix
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Suppose you want to obtain a transition matrix sending $N$ inputs into $M$ outputs.
You can then start with $N$ orthonormal vectors of length $MK$ with $K$ the dimension of the ancillary space (note that you can always think of such an orthonormal set of vectors as a subset of columns of some larger unitary matrix, like it is done in the question).
Denote ...
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With further understanding coming from the expanded question, I'm entirely revising my answer, but the original version is kept below in case it's useful.
The point, clearly, is to show how to simulate a probabilistic classical computation. So, we will store a classical distribution by using a diagonal density matrix:
$$
\rho=\sum_ip_i|i\rangle\langle i|.
$$...
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It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why $d^2$ dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are $d^2$ dimensional objects in the same sense that vectors in $\Bbb R^3$ are 3-dimensional (i.e., the cardinality of any basis set of $\...
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Yes, you can write a diagonally polarized state as a linear superposition of horizontally and vertically polarized states ("mixture" isn't the right term though; it's still a pure state). For instance, the $45^{\circ}$ diagonally polarized state $\lvert \nearrow \rangle$ may be expressed as
$$\lvert \nearrow \rangle = \frac{1}{\sqrt 2} \lvert \rightarrow \...
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One can perhaps guess the answer without full calculation. Noting that "tracing" intuitively means losing information, then, if you A is maximally entangled with B, then you lose information about B (or A) then you end up with no information about A. That is basically how qubits lose information to the environment (here B is like an environment for A).
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What does this density matrix say about the information that the players have about each others particles?
Nothing, if no further assumptions are made on the initial state. If $A$ and $B$ share a state $\rho$, a reduced state $\rho^A$ doesn't say anything about the information that $A$ has about $B$'s state.
Indeed, $B$ can have any state as far as $A$ ...
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The result is correct. You can see it in the other way. You have three two-qubit subsystems $A = \{1,2\}$, $B = \{3,4\}$ and $C = \{5,6\}$. The whole state is the product state on those three subsystems, i.e. the whole density matrix is $\rho_A \otimes \rho_B \otimes \rho_C$. Product state means there are absolutely no correlations between the states on ...
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Yes, the overall density matrix shared between Alice and Bob is $|\psi\rangle\langle\psi|$. To get the desnity matrix of either Alice or Bob, you should calculate
$$
\text{Tr}_B|\psi\rangle\langle\psi|\qquad\text{Tr}_A|\psi\rangle\langle\psi|
$$
respectively.
However, in this particular case, the calculation is much simply. Let $|\phi\rangle$ be the Bell ...
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In general, given two matrices $A$ and $B$ of dimensions $n_1\times n_2$ and $m_1\times m_2$, respectively, their tensor product $A\otimes B$ can be represented using the Kronecker product as
$$(A\otimes B)_{n_1 m_1,n_2m_2}=A_{n_1,n_2}B_{m_1, m_2}.$$
The indices on the left hand side are a standard way to enumerate the integers from $1$ to $n_1 m_1$ and from ...
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What you are looking for is https://en.wikipedia.org/wiki/Kronecker_product
Note that a column-vector can be considered as a matrix with the size $n \times 1$, so the Kronecker product rule also applies.
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