Density matrix of a qubit

  • Thread starter Dragonfall
  • Start date
  • #1
1,030
4

Main Question or Discussion Point

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

Answers and Replies

  • #2
410
0
In the appropriate basis,
[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
How is this a mixed state and not a pure state? Are some mixed states a pure state if you change the basis?
 
  • #4
410
0
It is mixed because it is obviously mixed =)
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
10,851
406
I'd like to learn these things better too. These are some of my thoughts:

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
0
The density matrix is always Hermitian, and it's trace is always 1. This is regardless of whether it represents a pure state or a mixed state. Once you diagonalize it, the condition of whether it is a pure state or a mixed state depends on whether one of the eigenvalues is 1 or not. If one is 1, and the others are zero, then it is a pure state. If more than one eigenvalue is non-zero, then it is a mixed state.

Remember what the density matrix represents.
 
  • #7
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
The density matrix is always Hermitian, and it's trace is always 1. This is regardless of whether it represents a pure state or a mixed state. Once you diagonalize it, the condition of whether it is a pure state or a mixed state depends on whether one of the eigenvalues is 1 or not. If one is 1, and the others are zero, then it is a pure state. If more than one eigenvalue is non-zero, then it is a mixed state.

Remember what the density matrix represents.
Of course. Thanks for clearing that up.
 
  • #8
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
Incidentally, there is a clear choice of basis that diagonalizes [itex]|\psi \rangle \langle \psi|[/itex]....
 
Last edited:
  • #9
1,030
4
I'm sorry but the Dirac notation and quantum mechanics is still unfamiliar to me so I'm going to ask some dumb questions:

[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
0
Watch out, [itex]
\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
0
The qubit can be in one of two states, or a superposition thereof. The two states are [itex]|\uparrow\rangle[/itex] and [itex]|\downarrow\rangle[/itex]. You can also think of these as column vectors [1,0] and [0,1].
 
  • #13
1,030
4
Ah, so it's another notation for |0> and |1>. What about my other questions?
 
  • #14
410
0
Looks like you edited your post right when I was writing my response, so anyway...
[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.
 
Last edited:
  • #16
410
0
I could've sworn I replied to this post. Anyway, what you have written there, [tex]D[/tex], is the most general 2x2 hermitian matrix, since those 4 matrices span the set of 2x2 hermitian matrices. It also has trace 1, since the pauli matrices are traceless. Therefore it is the most general density matrix.

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
But the coefficients in front of my expression has to be real, not complex. Do they still span the self-adjoint operators?
 
  • #18
410
0
Yes, the coefficients must be real in order to span self-adjoint operators. Well, you can take complex linear combinations and then impose Hermiticity to show that the coefficients have to be real.
 

Related Threads for: Density matrix of a qubit

Replies
1
Views
2K
Replies
2
Views
588
  • Last Post
Replies
6
Views
825
  • Last Post
Replies
4
Views
670
  • Last Post
Replies
3
Views
728
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
568
Replies
7
Views
971
Top