Bra-Ket Question: Constructing Langle Alpha & Beta

[Bobbo Snap]

Homework Statement

Consider a three-dimensional vector space spanned by an orthonormal basis |1\rangle, |2 \rangle, |3 \rangle. Kets |\alpha \rangle, |\beta \rangle are given by
|\alpha \rangle = i|1\rangle -2|2 \rangle -i|3\rangle, \qquad |\beta \rangle = i|1\rangle +2 |3\rangle.

part a) Construct \langle \alpha| \text{ and } \langle \beta | (in terms of the dual basis \langle 1|, \langle 2|, \langle 3|).

The Attempt at a Solution

I just want to check that I understand this correctly. Is the Bra the row vector that is basically the complex conjugate of the Ket, leading to the inner product? In this case,
\langle \alpha | = -i \langle 1 | -2 \langle 2| +i \langle 3| \qquad \langle \beta | = -i\langle 1| + 2 \langle 3|

 

[DrClaude]

Right.

 

[Bobbo Snap]

Thanks DrClaude. So if I have part a right,
\langle \alpha | = (-i, \, -2, \, i) \quad \text{ and } \quad \langle \beta | = (-i, \, 0, \, 2)

My calculation in the second part should be correct:
\langle \alpha | \beta \rangle = 1 + 2i \quad \text{ and } \quad \langle \beta | \alpha \rangle = 1 - 2i

Then the third part asks:
Find all nine matrix elements of the operator \hat{A} = |\alpha\rangle \langle \beta|, in this basis, and construct the matrix A. Is it hermitian?

How do I go about this? I don't see how to multiply |\alpha\rangle\langle \beta | to get nine elements.

 

[kevinferreira]

Then the third part asks:
Find all nine matrix elements of the operator \hat{A} = |\alpha\rangle \langle \beta|, in this basis, and construct the matrix A. Is it hermitian?

How do I go about this? I don't see how to multiply |\alpha\rangle\langle \beta | to get nine elements.

If |\alpha\rangle represents a column vector (3,1) and \langle \beta | a row vector (1,3) in terms of matrix multiplication what should (3,1)x(1,3) give you?

 

[HallsofIvy]

In vector terms the product <\alpha||\beta> is the "inner product"- after taking the complex conjugate, multiply corresponding terms and add. If |\alpha>= <a_1, a_2, a_3> and |\beta>= <b_1, b_2, b_3>, then <\alpha||\beta>= a_1b_1^*+ a_2b_2^*+ a_3b_3^*. ("*" is the complex conjugate.)

For the "exterior product", you form the nine products of every member of |\alpha> with every member of <\beta| as a matrix:
|\alpha><\beta|= \begin{bmatrix}a_1b_1^* & a_1b_2^* & a_1b_3^* \\ a_2b_1^* & a_2b_2^* & a_2b_3^* \\ a_3b_1^* & a_3b_2^* & a_3b_3^*\end{bmatrix}

 

[DrClaude]

Kevin and Ivy beat me to it. I will just stress that kets should be seen as column vectors, not row vectors as you wrote, and bras as their Hermitian conjugate.

 

[Bobbo Snap]

Thanks for the replies, I wasn't thinking of the Bra as a row vector. After doing the multiplication, I get
\hat{A} = | \alpha \rangle \langle \beta | = <br /> \begin{bmatrix} 1 &amp;0 &amp;2i\\ 2i &amp;0 &amp;-4\\ -1 &amp;0 &amp;-2i \end{bmatrix}
Which is not hermitian as \hat{A} \neq \hat{A}^\dagger. Correct?

 

[DrClaude]

Thanks for the replies, I wasn't thinking of the Bra as a row vector. After doing the multiplication, I get
\hat{A} = | \alpha \rangle \langle \beta | = <br /> \begin{bmatrix} 1 &amp;0 &amp;2i\\ 2i &amp;0 &amp;-4\\ -1 &amp;0 &amp;-2i \end{bmatrix}
Which is not hermitian as \hat{A} \neq \hat{A}^\dagger. Correct?

That looks correct.