Superposition of Hilbert space of qutrit states

[ma18]

Homework Statement

Given a orthonormal basis of the hilbert space of qutrit states: H = span (|0>, |1>, |2>)

write in abstract notation and also a chosen consistent matrix representation, the states

a) An equiprobable quantum superposition of the three elements of the basis

b) An equiprobable incoherent ensamble of the three possible elements of the basis

c) A bipartite state which is the tensor product of the two states built in a) and b)

The Attempt at a Solution

I got the following results but I am not sure if I did it correctly so I would love it it if someone could verify them or tell what what I did wrong

The chosen matrix representation is :

|0> = (1,0,0), |1> = (0,1,0), |2> = (0,0,1)

a) 1/sqrt(3) (|0>+ |1>+ |2>)

matrix form:

1/sqrt(3) (1,1,1)

b) 1/3 (|0><0|+ |1><1|+ |2><2|)

matrix form:

1/3 (1 0 0
0 1 0
0 0 1)

c)

1/3 sqrt(3)(1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1)I just used wikipedia to learn about the tensor product and I applied it to the 1*3 matrix from a and the 3*3 identity matrix (multiplied by 1/3) from b. I also don't know how to do c) with an abstract representation.

Any help would be much appreciated!

 

[DrClaude]

a) 1/sqrt(3) (|0>+ |1>+ |2>)

matrix form:

1/sqrt(3) (1,1,1)

That's not a matrix representation. My guess is that, even for pure states, you need to write the solution both in terms of the density operator and the density matrix.

 

[ma18]

That's not a matrix representation. My guess is that, even for pure states, you need to write the solution both in terms of the density operator and the density matrix.

Hi, thanks for the input.

It is annoying because there is no textbook for this course and the notes aren't the best.

Could you possibly help me a little more with how to formulate it?

In the notes the density matrix is said to be represented as: p = <ψ|ψ>

 

[DrClaude]

In the notes the density matrix is said to be represented as: p = <ψ|ψ>

That's actually
$$
\rho = \left| \psi \right\rangle \left\langle \psi \right|
$$
Using that, can you write down the density operator and the density matrix in part a?

 

[ma18]

That's actually
$$
\rho = \left| \psi \right\rangle \left\langle \psi \right|
$$
Using that, can you write down the density operator and the density matrix in part a?

Ah sorry, that's what I meant to write

So for part a) it would be

ρ=|ψ⟩⟨ψ| = 1/3 (|0>+ |1>+ |2>)(<0|+ <1|+ <2|) = 1/3(1+1+1) = 1 (Identity matrix?)

Huh but then b)

1/3 (|0><0|+ |1><1|+ |2><2|)

would also be equal to the identity if I assume that |0><0| and the like are equal to 1

I am a bit confused...

 

[DrClaude]

ρ=|ψ⟩⟨ψ| = 1/3 (|0>+ |1>+ |2>)(<0|+ <1|+ <2|) = 1/3(1+1+1) = 1 (Identity matrix?)

The part I've put in red is not correct. Note that something like $\left| 0 \right\rangle \left\langle 1 \right|$ cannot be simplified further.

To build the density matrix, you can calculate its elements as
$$
\rho_{ij} = \left\langle i \right| \hat{\rho} \left| j \right\rangle
$$
where $\hat{\rho}$ is the density operator.

 

[ma18]

The part I've put in red is not correct. Note that something like $\left| 0 \right\rangle \left\langle 1 \right|$ cannot be simplified further.

To build the density matrix, you can calculate its elements as
$$
\rho_{ij} = \left\langle i \right| \hat{\rho} \left| j \right\rangle
$$
where $\hat{\rho}$ is the density operator.

Okay, so then I get

p = 1/3(|0><0|+|0><1|+|0><2|+|1><0|+|1><1|+|1><2|+|2><0|+|2><1|+|2><2|)

and then constructing the matrix, it becomes

p = 1/3 (1 1 1
1 1 1
1 1 1)

 

[ma18]

And then for b)

p = 1/9 = (|0><0|+|1><1|+|2><2|)(<0|0>+<1|1>+<2|2>) = 1/3 (|0><0|+|1><1|+|2><2|)
= 1/3 (1 1 1
1 1 1
1 1 1)

which is the same result as in a)

Is this correct?

 

[DrClaude]

And then for b)

p = 1/9 = (|0><0|+|1><1|+|2><2|)(<0|0>+<1|1>+<2|2>) = 1/3 (|0><0|+|1><1|+|2><2|)

I don't understand what you did in the middle. The problem mentions and "incoherent ensemble" (the technical term is mixed state), which means that you can't write it as a ket. You can only write it directly as a density operator, which is what you did in the OP.

= 1/3 (1 1 1
1 1 1
1 1 1)

which is the same result as in a)

Is this correct?

This is not correct. Again, this you wrote correctly in the OP.

 

[ma18]

I don't understand what you did in the middle. The problem mentions and "incoherent ensemble" (the technical term is mixed state), which means that you can't write it as a ket. You can only write it directly as a density operator, which is what you did in the OP.This is not correct. Again, this you wrote correctly in the OP.

Ah, now I'm confused. I thought you said that to construct the density matrix I needed to do

P = |phi><phi|

so I did that in the middle step (there shouldn't be an equal sign after 1/9) and then I did <i|P|j> for the terms in the matrix.

So if the original way I put it is already as the density matrix then it would be

(1 1 1
1 1 1
1 1 1)

instead?

Is what I did for the part a) matrix representation correct?

 

[DrClaude]

What you did for part a is correct.

For part b, the problem is that you can't write a ket for a mixed state. In other words, there is no state vector $\left| \psi \right\rangle$ that can describe the state of the system. This is precisely why the density operator is needed; otherwise, we could stick to using kets.

When the system is in a pure state, it can be described by a state vector $\left| \psi \right\rangle$, and the corresponding density operator is
$$
\hat{\rho} = \left| \psi \right\rangle \left\langle \psi \right|
$$
Where the system is in a mixed state, the density operator is built from a basis of states $\left| \phi_n \right\rangle$ as
$$
\hat{\rho} = \sum_n p_n \left| \phi_n \right\rangle \left\langle \phi_n \right|
$$
where $p_n$ is the probability of finding the system in state $\left| \phi_n \right\rangle$.

To come back to your exercise, for an equiprobable incoherent ensemble of the three possible elements of the basis, you thus get
$$
\hat{\rho} = \frac{1}{3} \left| 0 \right\rangle \left\langle 0 \right| + \frac{1}{3} \left| 1 \right\rangle \left\langle 1 \right| + \frac{1}{3} \left| 2 \right\rangle \left\langle 2 \right|
$$
In matrix representation, using $\rho_{ij} = \left\langle i \right| \hat\rho \left| j \right\rangle$, you get
$$
\rho = \frac{1}{3} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix}
$$

 

[ma18]

What you did for part a is correct.

For part b, the problem is that you can't write a ket for a mixed state. In other words, there is no state vector $\left| \psi \right\rangle$ that can describe the state of the system. This is precisely why the density operator is needed; otherwise, we could stick to using kets.

When the system is in a pure state, it can be described by a state vector $\left| \psi \right\rangle$, and the corresponding density operator is
$$
\hat{\rho} = \left| \psi \right\rangle \left\langle \psi \right|
$$
Where the system is in a mixed state, the density operator is built from a basis of states $\left| \phi_n \right\rangle$ as
$$
\hat{\rho} = \sum_n p_n \left| \phi_n \right\rangle \left\langle \phi_n \right|
$$
where $p_n$ is the probability of finding the system in state $\left| \phi_n \right\rangle$.

To come back to your exercise, for an equiprobable incoherent ensemble of the three possible elements of the basis, you thus get
$$
\hat{\rho} = \frac{1}{3} \left| 0 \right\rangle \left\langle 0 \right| + \frac{1}{3} \left| 1 \right\rangle \left\langle 1 \right| + \frac{1}{3} \left| 2 \right\rangle \left\langle 2 \right|
$$
In matrix representation, using $\rho_{ij} = \left\langle i \right| \hat\rho \left| j \right\rangle$, you get
$$
\rho = \frac{1}{3} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix}
$$

Ah, I see. That makes sense.

Thank you for you help.

For the tensor product of the states in a) and b),the matrix representation would just be a 9x9 matrix but how would it be represented in abstract notation?