if only I knew what the hell a density matrix means, physically. Not how to define it, what it means. This information seems to have been left out of physics textbooks and Wikipedia.
I’m here to help!
Recall that with N qubits we have a 2^N dimensional space whose basis vectors correspond to all possible bit strings of length N. A pure state, like ( |00> + |11> ) / √2, is a vector in that space. When we measure a pure state, the probability of each bit string is the squared dot product of the state vector with the corresponding basis vector. We can assume that the state vector is a unit vector, which makes its squared dot products with all basis vectors (i.e. all probabilities) sum to 1. We can also rotate the vector, and probabilities will still sum to 1, because length is invariant under rotation. Math is wonderful.
On the other hand, a mixed state like “|0> or |1> with probability 1⁄2 each” is a norm on that space. When we measure a mixed state, the probability of each bit string is that norm applied to the corresponding basis vector. Each pure state corresponds to a degenerate norm which does squared projection onto the state vector, so you can think of it as a straight line; whereas most mixed states correspond to non-degenerate norms, so you can think of them as ellipsoids. We can assume that the norm has trace 1, so the norms of all basis vectors (i.e. all probabilities) sum to 1. We can also rotate the norm, or equivalently rotate the basis, and probabilities will still sum to 1, because trace is invariant under rotation. Math is wonderful.
That fixes a small annoyance I noticed this morning. There’s no way to distinguish |0> from -|0> by rotation, negation and measurement, so they should be the same. As vectors they are different, but as norms (represented by 2^N x 2^N matrices, known as “density matrices”) they are indeed the same! The correspondence between mixed states and density matrices is one-to-one.
The computational rules are very simple:
For a basis state, the density matrix has one 1 on the diagonal and 0 everywhere else. For example, the pure state |0> has density matrix ((1,0),(0,0)).
For a pure state that’s a superposition of basis states, you take the superposition coefficients a_i for i from 1 to 2^N, and build the matrix {a_i*a_j}. For example, the pure state ( |0> + |1> ) / √2 has superposition coefficients (1/√2, 1/√2), so the density matrix is ((1/2,1/2),(1/2,1/2)).
For a mixed state that’s a probabilistic mixture of pure states, you just take a weighted sum of matrices. For example, the mixed state “|0> or |1> with probability 1⁄2 each” has density matrix ((1/2,0),(0,1/2)). Note that this matrix, unlike the other two, isn’t degenerate—it’s a Euclidean norm. Now it’s obvious why this mixed state is invariant under rotation, and why no pure state can do the same, which we figured out the hard way in one of the previous comments.
No. The property which you are describing is not “mixedness” (technical term: “purity”). That the state vector in question can’t be written as a tensor product of state vectors makes it an *entangled* state.
Mixed states are states which cannot be represented by *any* state vector. You need a density matrix in order to write them down.
(5/?)
Eliezer asked long ago:
I’m here to help!
Recall that with N qubits we have a 2^N dimensional space whose basis vectors correspond to all possible bit strings of length N. A pure state, like ( |00> + |11> ) / √2, is a vector in that space. When we measure a pure state, the probability of each bit string is the squared dot product of the state vector with the corresponding basis vector. We can assume that the state vector is a unit vector, which makes its squared dot products with all basis vectors (i.e. all probabilities) sum to 1. We can also rotate the vector, and probabilities will still sum to 1, because length is invariant under rotation. Math is wonderful.
On the other hand, a mixed state like “|0> or |1> with probability 1⁄2 each” is a norm on that space. When we measure a mixed state, the probability of each bit string is that norm applied to the corresponding basis vector. Each pure state corresponds to a degenerate norm which does squared projection onto the state vector, so you can think of it as a straight line; whereas most mixed states correspond to non-degenerate norms, so you can think of them as ellipsoids. We can assume that the norm has trace 1, so the norms of all basis vectors (i.e. all probabilities) sum to 1. We can also rotate the norm, or equivalently rotate the basis, and probabilities will still sum to 1, because trace is invariant under rotation. Math is wonderful.
That fixes a small annoyance I noticed this morning. There’s no way to distinguish |0> from -|0> by rotation, negation and measurement, so they should be the same. As vectors they are different, but as norms (represented by 2^N x 2^N matrices, known as “density matrices”) they are indeed the same! The correspondence between mixed states and density matrices is one-to-one.
The computational rules are very simple:
For a basis state, the density matrix has one 1 on the diagonal and 0 everywhere else. For example, the pure state |0> has density matrix ((1,0),(0,0)).
For a pure state that’s a superposition of basis states, you take the superposition coefficients a_i for i from 1 to 2^N, and build the matrix {a_i*a_j}. For example, the pure state ( |0> + |1> ) / √2 has superposition coefficients (1/√2, 1/√2), so the density matrix is ((1/2,1/2),(1/2,1/2)).
For a mixed state that’s a probabilistic mixture of pure states, you just take a weighted sum of matrices. For example, the mixed state “|0> or |1> with probability 1⁄2 each” has density matrix ((1/2,0),(0,1/2)). Note that this matrix, unlike the other two, isn’t degenerate—it’s a Euclidean norm. Now it’s obvious why this mixed state is invariant under rotation, and why no pure state can do the same, which we figured out the hard way in one of the previous comments.
Do you by any chance have a typo here? Sorry if I am wrong, since I don’t actually know quantum information theory.
I think this state is mixed, since it’s a sum of two vectors, which can’t be represented as just one kronecker product.
No. The property which you are describing is not “mixedness” (technical term: “purity”). That the state vector in question can’t be written as a tensor product of state vectors makes it an *entangled* state.
Mixed states are states which cannot be represented by *any* state vector. You need a density matrix in order to write them down.