- #1

- 1,030

- 4

What is the arbitrary density matrix of a mixed state qubit?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Dragonfall
- Start date

- #1

- 1,030

- 4

What is the arbitrary density matrix of a mixed state qubit?

- #2

- 410

- 3

[tex]\rho = \alpha \left| \uparrow \right\rangle \left\langle \uparrow \right| + \beta \left| \downarrow \right\rangle \left\langle \downarrow \right|[/tex],

where [tex]\alpha + \beta = 1[/tex]

- #3

- 1,030

- 4

- #4

- 410

- 3

A pure state is of the form [tex]\left| \psi \right\rangle \left\langle \psi \right|[/tex], and you can't write the state given in that way. It is most general because you can always diagonalize a mixed state.

In order to purify a mixed state you need to enlarge your system, iirc.

- #5

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,872

- 419

Take an arbitrary superposition [itex]\left|\psi\rangle=a\left|\uparrow\rangle+b\left|\downarrow\rangle[/itex] and consider its pure-state density matrix:

[tex]\rho=\left|\psi\rangle\langle\psi\right|=|a|^2\left|\uparrow\rangle\langle\uparrow\right| +ab^*\left|\uparrow\rangle\langle\downarrow\right| +a^*b\left|\downarrow\rangle\langle\uparrow\right| +|b|^2\left|\downarrow\rangle\langle\downarrow\right|[/tex]

A different choice of basis would have made it diagonal:

[tex]\left|\psi\rangle\langle\psi\right|=\alpha\left|\uparrow\rangle\langle\uparrow\right| +\beta\left|\downarrow\rangle\langle\downarrow\right|[/tex]

Lbrits, this looks a lot like your "obviously mixed" state, so if you're right, then it's the condition [itex]\alpha+\beta=1[/itex] that makes this a mixed state, and that doesn't seem obvious at all.

I would have guessed that the general mixed density matrix is

[tex]a\left|\uparrow\rangle\langle\uparrow\right| +b\left|\uparrow\rangle\langle\downarrow\right| +c\left|\downarrow\rangle\langle\uparrow\right| +d\left|\downarrow\rangle\langle\downarrow\right|[/tex]

with a,b,c,d arbitrary complex numbers, and that this represents a pure state if and only if a and d are both real and b=c*. But in that case, assuming lbrits is right, then [itex]\alpha[/itex] and [itex]\beta[/itex] in the general mixed state would have to have non-zero and opposite imaginary parts. That seems like a strange condition.

- #6

- 410

- 3

Remember what the density matrix represents.

- #7

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,872

- 419

Of course. Thanks for clearing that up.

Remember what the density matrix represents.

- #8

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,950

- 19

Incidentally, there is a clear choice of basis that diagonalizes [itex]|\psi \rangle \langle \psi|[/itex]....

Last edited:

- #9

- 1,030

- 4

[itex]\left|\psi\right>\left<\psi\right|[/itex] is the projector onto the space spanned by [itex]\left|\psi\right>[/itex], correct? So a if we change the basis to one containing [itex]\left|\psi\right>[/itex] then [itex]\left|\psi\right>[/itex] is an eigenvector with eigenvalue 1?

Now given an arbitrary density matrix, it is self-adjoint so we can uniquely diagonize it. Once we do so we can find out whether the state it describes is pure?

So ONE POSSIBLE density matrix of an arbitrary qubit is [itex]\alpha\left| 0\right>\left< 0\right|+\beta\left| 1\right>\left< 1\right|[/itex]?

- #10

- 410

- 3

\alpha\left| 0\right>+\beta\left| 1\right>

[/itex] is a state, not an operator.

Think of [itex]\left|v\right\rangle[/itex] as a vector [itex]\vec{v}[/itex]. Then the expression [itex]\left|v \right\rangle \left\langle w \right|[/itex] represents the matrix [itex]\vec{v}^T \vec{w}[/itex]. If you take the operator and multiply on the left by a bra, and on the right by a ket, you see that it acts like an object with two indices. I hope this is clear. I guess the Dirac notation makes it easy to abstract away the whole matrix/index business, but it is handy to understand both pictures.

- #11

- 1,030

- 4

What is [itex]\left|\uparrow\rangle[/itex]?

- #12

- 410

- 3

- #13

- 1,030

- 4

Ah, so it's another notation for |0> and |1>. What about my other questions?

- #14

- 410

- 3

[itex]

\alpha\left| 0\right>\left< 0\right|+\beta\left| 1\right>\left< 1\right|

[/itex]

Is actually the most general, because any 2x2 density matrix can be brought to this form by a unitary transformation. It depends on whether you regard density matrices related by unitary transformations as somehow distinct or not.

If the density matrix has more than one non-zero eigenvalue, then it is not pure. It is mixed. If it has a single non-zero eigenvalue, it is pure.

- #15

- 1,030

- 4

I'm trying to show that an arbitrary density matrix D for a qubit can be written as

[tex]D=\frac{I+\vec{r}\cdot\vec{\rho}}{2}[/tex].

Where [itex]\vec{r}[/itex] is a real 3 dimensional vector of length less than or equal to 1, and [itex]\vec{\rho}[/itex] is the 3 Pauli matrices excluding identity.

But if an arbitrary density matrix is [itex]

\alpha\left| 0\right>\left< 0\right|+\beta\left| 1\right>\left< 1\right|

[/itex] then it wouldn't work.

[tex]D=\frac{I+\vec{r}\cdot\vec{\rho}}{2}[/tex].

Where [itex]\vec{r}[/itex] is a real 3 dimensional vector of length less than or equal to 1, and [itex]\vec{\rho}[/itex] is the 3 Pauli matrices excluding identity.

But if an arbitrary density matrix is [itex]

\alpha\left| 0\right>\left< 0\right|+\beta\left| 1\right>\left< 1\right|

[/itex] then it wouldn't work.

Last edited:

- #16

- 410

- 3

In order to diagonalize a matrix, you have to change basis. So, in a particular basis, D is the most general, since in order to make it diagonal, you would have to go into a different basis. Hope that is clear.

- #17

- 1,030

- 4

- #18

- 410

- 3

Share: