Ket-vector and bra-vector in dual space (for qutrits)

[limarodessa]

Homework Statement

There are three qutrits which are connected by quantum entanglement:

$|\Psi\rangle=\frac{1}{\sqrt{3}}\left( {|021\rangle+|102 \rangle+ |102 \rangle} \right)$

How can I to describe $\langle\Psi|$ ?

$\langle\Psi|=\frac{1}{\sqrt{3}}\left( {\langle ... |+\langle ... | + \langle ... | } \right)$

What must I write instead points in angle brackets - ${\langle ... |}$ ?

Homework Equations

The Attempt at a Solution

I think in case:

$|\Psi\rangle=\frac{1}{\sqrt{3}}\left( {|012\rangle+|120 \rangle+ |201 \rangle} \right)$

it will :

$\langle\Psi|=\frac{1}{\sqrt{3}}\left( {\langle 210 |+\langle 021 | + \langle 102 | } \right)$

but i do not know what I must write in case

$|\Psi\rangle=\frac{1}{\sqrt{3}}\left( {|021\rangle+|102 \rangle+ |102 \rangle} \right)$

- I do not know what it will in
$\langle\Psi|=\frac{1}{\sqrt{3}}\left( {\langle ... |+\langle ... | + \langle ... | } \right)$

 

[fzero]

I think in case:

$|\Psi\rangle=\frac{1}{\sqrt{3}}\left( {|012\rangle+|120 \rangle+ |201 \rangle} \right)$

it will :

$\langle\Psi|=\frac{1}{\sqrt{3}}\left( {\langle 210 |+\langle 021 | + \langle 102 | } \right)$

If try to use this to compute $\langle \Psi | \Psi \rangle$, you'll find zero, which is not possible. Therefore the expression for $\langle \Psi|$ is wrong.

In order to understand why, you should make sure that you understand exactly what we mean by the expression $|abc\rangle$. This is a state in a Hilbert space describes the composite states of 3 identical particles, each of which can be in 3 orthonormal single-particle states $| a \rangle$, $a-0,1,2$. In fact,

$$ | a b c \rangle = |a\rangle \otimes | b\rangle \otimes | c\rangle,$$

where the $\otimes$ symbol denotes a tensor product. The first term in the tensor product is the state of the first particle, the second term for the second particle, etc.

Since the single particle states can be taken to satisfy an orthonormality condition $\langle a' | a \rangle = \delta_{a'a}$, we want the conjugate states to satisfy the analogous condition

$$ \langle a'b'c' | a bc\rangle = \delta_{a'a}\delta_{b'b}\delta_{c'c}.$$

You should try to work out the form of $\langle a'b'c' |$ such that this is true. If necessary, consider the general form that the bra could take:

$$\langle a'b'c' | = \sum_{abc} M_{a'b'c'abc} \langle a | \otimes \langle b|\otimes \langle c|.$$