A little notation help, on quantum coding

1. May 18, 2009

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 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 :)

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2. May 18, 2009

Cyosis

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: May 18, 2009
3. May 18, 2009

monica1977

aint <v2 is the transpose of v2> ???

4. May 18, 2009

Cyosis

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?

5. May 18, 2009

dx

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: May 18, 2009
6. May 18, 2009

monica1977

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: May 18, 2009
7. May 18, 2009

Cyosis

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:
$$\frac{1}{2}\left( \begin{matrix} i*-i & i*-1 \\ 1*-i & 1*-1 \end{matrix}\right)= \frac{1}{2}\left( \begin{matrix} 1 & -i \\ -i & -1 \end{matrix}\right)$$

8. May 18, 2009

monica1977

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: May 18, 2009
9. May 18, 2009

monica1977

Any one ?

10. May 18, 2009

Cyosis

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.

11. May 18, 2009

diazona

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.