# A little notation help, on quantum coding

• monica1977
In summary, the expression |v_1\rangle \langle v_2| represents the conjugate transpose of |v_2 \rangle.

#### monica1977

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 don't understand how it forms into another matrix. I have the answer attached as well , but could some one explain ? (I don't think the matrix A is needed , i just copied it as well )

Cheers for any help guys :)

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

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aint <v2 is the transpose of v2> ?

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?

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

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

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Cheers Cyosis I understand that better now , but i still don't 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 ?

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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 ?

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Any one ?

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.

Cyosis said:
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.