# A little notation help, on quantum coding

## Main Question or Discussion Point

Hi, I wanted to know how to solve this question , its not a homework question i am really asking for , more the general way to solve these types of questions.......... I dont understand how it forms into another matrix. I have the answer attached aswell , but could some one explain ? (I dont think the matrix A is needed , i just copied it aswell )

Cheers for any help guys :)

#### Attachments

• 8.9 KB Views: 286
• 1.7 KB Views: 257

Related Quantum Physics News on Phys.org
Cyosis
Homework Helper
Do you know what $\langle v_2|$ represents? It is the conjugate transpose of $|v_2 \rangle$. So $\langle v_2|=\frac{1}{\sqrt{2}}(-i,-1)$.

Therefore
$$|v_1 \rangle \langle v_2|=\frac{1}{2} \binom{i}{1}(-i,-1)$$

Last edited:
aint <v2 is the transpose of v2> ???

Cyosis
Homework Helper
Almost, we are working with complex vectors here so it is the conjugate transpose. Do you know how to form a matrix from the the expression in my previous post?

dx
Homework Helper
Gold Member
To represent an operator as a matrix, you must choose a basis. In this case, the question presumably wants you to write it in the same basis that $$|v_1\rangle$$ and $$|v_2\rangle$$ are given in. You are given that

$$|v_1\rangle = \frac{1}{\sqrt{2}} ( i|1\rangle + |2\rangle)$$

$$|v_2\rangle = \frac{1}{\sqrt{2}} (i|1\rangle -|2\rangle)$$

The operator is

$$|v_1\rangle\langle v_2|$$.

The matrix elements of this operator in the basis $$|1\rangle, |2\rangle$$ are

$$A_{ij} = \langle i|v_1\rangle\langle v_2|j\rangle$$.

For example, the (1,1) element of the matrix will be

$$A_{11} = \langle 1|v_1\rangle\langle v_2|1\rangle = \langle 1|v_1\rangle \langle 1 | v_2 \rangle ^{*} = \frac{i}{\sqrt{2}}\frac{-i}{\sqrt{2}} = \frac{1}{2}$$.

Last edited:
Cheers Cyosis I understand that better now , but i still dont understand how what you wrote is i matrix and not a scalar ? , can u show me for a general case ? , or just show me how the answer is acheived ?

Last edited:
Cyosis
Homework Helper
To calculate $|v_1 \rangle \langle v_2|=\frac{1}{2} \binom{i}{1}(-i,-1)$ just think of it as two matrices. You multiply the first row of v1 (i) with the first column of v2 (-i).

What you will get is this:
$$\frac{1}{2}\left( \begin{matrix} i*-i & i*-1 \\ 1*-i & 1*-1 \end{matrix}\right)= \frac{1}{2}\left( \begin{matrix} 1 & -i \\ -i & -1 \end{matrix}\right)$$

Hi I just wanted to ask quickly , does this only apply to only complex vectors ? , what if they was all real values ? , is it still the same then ????

Last edited:
Any one ?

Cyosis
Homework Helper
Yes it's the same. The only difference is that you usually take the normal transpose for real valued vectors since there is nothing to conjugate.

diazona
Homework Helper
Yes it's the same. The only difference is that you usually take the normal transpose for real valued vectors since there is nothing to conjugate.
Or the way I like to think about it, the conjugate transpose for real vectors (or matrices) is the normal transpose, since the conjugate of a real number is just the same number.