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$\renewcommand{ket}[1]{|#1\rangle}$ I am facing difficulty in understanding how the right hand side is coming in equation A below

In $H$ of dimention 4, the vector $$ \sqrt{\frac{2}{3}} \ket{01} + \frac{i}{\sqrt{3}}\ket{11} = \sqrt{\frac{2}{3}}\ket{0} \otimes \ket{1} + \frac{i}{\sqrt{3}}\ket{1} \otimes \ket{1} \tag{A} $$ in Diract notation can be alternatively written as the column matrix $$ \left( \begin{array}{c} 0 \\ \sqrt{\frac{2}{3}} \\ 0 \\ \frac{i}{\sqrt{3}} \end{array} \right) \, . \tag{B} $$

Also how it is being represented in B?

What I understand is basis vectors here are $\ket{01}$ and $\ket{11}$, similar to $i$, and $j$ unit vectors. Since this is 4 dimensional space, I must see 4 basis vectors in A, but it is not there, why? If I understand this I will put the coefficients of these basis vectors in a column and will understand how B comes .

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  • $\begingroup$ $|00\rangle, |01\rangle, |10\rangle, |11\rangle$. $\endgroup$
    – Meng Cheng
    Jun 17, 2015 at 5:31
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    $\begingroup$ Please do not post images of equations. Type the equations you need. If you hit the "edit" button on your question you can see how I did the equation formatting. $\endgroup$
    – DanielSank
    Jun 17, 2015 at 5:34
  • $\begingroup$ @DanielSank Thanks for editing. I will keep this in mind in all future posts. $\endgroup$
    – gpuguy
    Jun 17, 2015 at 5:41
  • $\begingroup$ @MengCheng I dont see these basis vectors in equation A, why so ? $\endgroup$
    – gpuguy
    Jun 17, 2015 at 5:42
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    $\begingroup$ Some of the coefficients are zero, as evident in B. $\endgroup$
    – Omry
    Jun 17, 2015 at 6:01

2 Answers 2

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Here taking $|0\rangle$ and $|1\rangle$ as orthonormal basis for 2 dimensional hilbert space. Now $|00\rangle ,|01\rangle,|10\rangle,|11\rangle$ are orthogonal to each other ( take the inner product of any two it will be zero, eg. $\langle 00|01 \rangle= \langle0|0\rangle \langle 0|1\rangle =0 $ ). Thus any vector of a 4 dimensional dimensional hilbert space can be represented in linear combination of these 4 vectors ( basis vectors ) as they are orthogonal to each other. That is any vector in 4 dimensional hilbert space can be given as $|v\rangle = \sum_{i,j=0,1}\alpha_{ij}|ij\rangle$ where $\alpha_{ij}$ are some complex numbers. But this does not mean that each vector has to have some component along each of the four basis vectors. That is what the case is in your example, meaning $\alpha_{ij}$ can take 0 also as values. In your example $\alpha_{00}=\alpha_{10}=0$, you can always write your given vector as $0.|00\rangle + \sqrt{\frac{2}{3}}|01\rangle +0.|10\rangle+\frac{i}{\sqrt{3}}|11\rangle$. Now $0.|v\rangle$ for some vector $|v\rangle$ is null vector and $|a\rangle+ |\phi\rangle = |a\rangle$ for $|a\rangle$ be any vector and $|\phi\rangle$ being null vector, that is why you don't need to write them in your expression. I hope I answered your question.
EDIT
You should look up tensor product and their properties. Still one can understand tensor product in terms of column matrices. Say $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ taking tensor product $|0\rangle \otimes |1\rangle= |01\rangle= \begin{pmatrix} 1* \begin{pmatrix} 0 \\ 1 \end{pmatrix}\\ 0* \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{pmatrix}= \begin{pmatrix} 0 \\ 1 \\ 0 \\0 \end{pmatrix}$
Similarly it can be found out for other terms and in general.

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  • $\begingroup$ What is the symbol with x in a circle on right hand side in (A)? How is it coming ? $\endgroup$
    – gpuguy
    Jun 17, 2015 at 6:39
  • $\begingroup$ that is tensor product , it is equivalent to write $|ij\rangle=|i\rangle \otimes |j\rangle$ $\endgroup$ Jun 17, 2015 at 6:40
  • $\begingroup$ @gpuguy I made an EDIT hope it helps. $\endgroup$ Jun 17, 2015 at 6:56
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$\newcommand{\ket}[1]{\left| #1 \right>}$Note that you can write a tensor product as a matrix in the following way:

$$A\otimes B = \begin{pmatrix} A_{11}B & \ldots & A_{1m}B\\ \vdots & \ddots & \vdots\\ A_{m1}B & \ldots & A_{mm} B \end{pmatrix}$$

where $A$ is a $m\times m$ matrix and $B$ is a $n\times n$ matrix. Notice that the resulting matrix is a $nm\times nm$ matrix. For the case of 2d vectors you can use the analogy and write:

$$\ket+ \otimes \ket+ = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\end{pmatrix} \quad \ket+ \otimes \ket- = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\end{pmatrix}$$ $$\ket- \otimes \ket+= \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix}\quad \ket- \otimes \ket- = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}$$

In general a vector can be written as the superposition of these basis vectors:

$$\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix} = d\ket{++} +c\ket{+-} +b\ket{-+} +d\ket{--} $$

In your case some of these coefficients happen to be zero ie

$$\left( \begin{array}{c} 0 \\ \sqrt{\frac{2}{3}} \\ 0 \\ \frac{i}{\sqrt{3}} \end{array} \right) = \frac{i}{\sqrt{3}} \ket{++} + 0\ket{+-} + \sqrt{\frac{2}{3}} \ket{-+} + 0 \ket{--}$$

Notice that I have used $\ket+ = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\ket - = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ because of your notation in the question, which is somewhat unusual.

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