Density matrix for bell states

1. Mar 1, 2013

Hi
I have three states (I believe bell states) and want to find the density matrix, am I right in thinking:
1) $\frac{|00> + |11>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} \\ \end{array} \right)$ (because it is pure)

2) $\frac{|00> - |11>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} \\ \end{array} \right)$ (because it is pure)

3) $\frac{|01> + |01>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)$ (because it is mixed)

2. Mar 2, 2013

spookyfw

Shouldn't the density matrix be 4x4? The first one should be something like

$$\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

The others then analogously, but a 100% I am not certain, because looking at the Schmidt-coefficients, they are not all the same (2 are actually 0) as required for a maximally entangled state. Anyone can solve this issue?

3. Mar 2, 2013

kith

You get the density matrix of the composite system by writing out ρ= |ψ><ψ| for your states. This is a 4x4 matrix.

If you want to get the reduced density matrix of one subsystem,you have to trace over the other system. You then get a 2x2 matrix.

PS: Why do you think that the third state is not pure? Every ket corresponds to a pure state. You seem to have some fundamental misunderstanding. You need to give more details about your calculations.

Last edited: Mar 2, 2013
4. Mar 2, 2013

kith

You need only as many coefficents as the dimension of the Hilbert space of the smaller subsystem. Here, we have two 2-dimensional Hilbert spaces, so we need two coefficients.

5. Mar 2, 2013

Please could you describe the difference between a pure and a mixed state?
I am also unclear as to how to calculate the density matrix, could you give me a simple example, I can't see why $\sum_i |\psi_i><\psi_i|$ would give a 4x4 matrix.

Would you have to decompose $|00>$ into |0>|0> and then into,
(A|0> + B|1>)Tensor-product(C|0>+D|1>)? I am a bit confused

6. Mar 3, 2013

tom.stoer

A pure state can be a linear combination of other pure states. This is the quantum mechanical superposition.

For a pure state |ψ> we simply chose this state |ψ> to construct the density matrix

ρ = |ψ><ψ|

As you can see you have ρ2 = ρ, so ρ is a projector (here we do not care whether there are other states, e.g. u and v, which can be combined via a linear superposition to get ψ).

A mixed state is always described via a density matrix which is not a projector.

For a mixed state, in a basis where the density matrix is diagonal, you have two or more states with non-vanishing probability. So for two orthogonal states contributing to a density matrix you have

ρ = p|u><u| + (1-p)|v><v|

Note that for a diagonal density matrix the trace of the probailities must be = 1.

Testing whether ρ is a projector you find

ρ2 = p2|u><u| + (1-p)2|v><v| ≠ ρ

The probabilities p and (1-p) in ρ are classical probabilities, not related to the quantum mechanical superpositions. So if you would construct ψ from u and v using quantum mechanical superpositions you get the pure state as described above.

7. Mar 3, 2013

kith

Have you calculated $\rho = |\psi><\psi| = \frac{1}{2}(|00> + |11>)(<00|+<11|)$? (First example from you)

ρ has 16 entries which are the coefficients in front of the 16 possible combinations |ij><kl|. Most of them are zero, but 4 are non-zero. If you arrange them in matrix form, you are done.

No decomposing of |ij> is required, because we are now dealing with the combined state space of two quibits. There are states which can't be decomposed and these are the entangled states.

Last edited: Mar 3, 2013