# 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

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.