A little notation help, on quantum coding

  • Thread starter monica1977
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  • #1

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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
Cyosis
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Do you know what [itex]\langle v_2|[/itex] represents? It is the conjugate transpose of [itex]|v_2 \rangle[/itex]. So [itex]\langle v_2|=\frac{1}{\sqrt{2}}(-i,-1)[/itex].

Therefore
[tex]
|v_1 \rangle \langle v_2|=\frac{1}{2} \binom{i}{1}(-i,-1)
[/tex]
 
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  • #3
aint <v2 is the transpose of v2> ???
 
  • #4
Cyosis
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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
dx
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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 [tex] |v_1\rangle [/tex] and [tex] |v_2\rangle [/tex] are given in. You are given that

[tex]|v_1\rangle = \frac{1}{\sqrt{2}} ( i|1\rangle + |2\rangle) [/tex]

[tex]|v_2\rangle = \frac{1}{\sqrt{2}} (i|1\rangle -|2\rangle) [/tex]

The operator is

[tex] |v_1\rangle\langle v_2| [/tex].

The matrix elements of this operator in the basis [tex] |1\rangle, |2\rangle [/tex] are

[tex] A_{ij} = \langle i|v_1\rangle\langle v_2|j\rangle [/tex].

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

[tex] 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} [/tex].
 
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  • #6
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 ?
 
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  • #7
Cyosis
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To calculate [itex]|v_1 \rangle \langle v_2|=\frac{1}{2} \binom{i}{1}(-i,-1)[/itex] 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:
[tex]\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)[/tex]
 
  • #8
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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  • #9
Any one ?
 
  • #10
Cyosis
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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
diazona
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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.
 

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