Basic Bra-Ket Question

1. Feb 17, 2013

Bobbo Snap

1. The problem statement, all variables and given/known data

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 $\langle 1|, \langle 2|, \langle 3|$).

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,
$$\langle \alpha | = -i \langle 1 | -2 \langle 2| +i \langle 3| \qquad \langle \beta | = -i\langle 1| + 2 \langle 3|$$

Last edited: Feb 17, 2013
2. Feb 18, 2013

Staff: Mentor

Right.

3. Feb 18, 2013

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:
$$\langle \alpha | \beta \rangle = 1 + 2i \quad \text{ and } \quad \langle \beta | \alpha \rangle = 1 - 2i$$

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.

4. Feb 18, 2013

kevinferreira

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?

5. Feb 18, 2013

HallsofIvy

Staff Emeritus
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:
$$|\alpha><\beta|= \begin{bmatrix}a_1b_1^* & a_1b_2^* & a_1b_3^* \\ a_2b_1^* & a_2b_2^* & a_2b_3^* \\ a_3b_1^* & a_3b_2^* & a_3b_3^*\end{bmatrix}$$

6. Feb 18, 2013

Staff: Mentor

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.

7. Feb 18, 2013

Bobbo Snap

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

8. Feb 19, 2013

Staff: Mentor

That looks correct.