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The issue is

1.which one is right? \begin{align} & (|{{w}_{1}}...{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}}...{{w}_{n}}|? \\ & (|{{w}_{1}}...{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{n}}...{{w}_{1}}|? \\ & (|{{w}_{1}}{{U}_{1}}{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}}U_{1}^{\dagger }{{w}_{n}}|? \\ & (|{{w}_{1}}{{U}_{1}}{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{n}}U_{1}^{\dagger }{{w}_{1}}|? \\ \end{align}

2.judge whether it is correct \begin{align} & \langle w_{2}^{{}}w_{1}^{{}}\text{ }\!\!|\!\!\text{ }{{w}_{1}}{{U}_{q2}}{{w}_{2}}\rangle \text{?=}\langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }{{U}_{q2}}\text{ }\!\!|\!\!\text{ }{{w}_{2}}\rangle \\ & \to \langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }\otimes \langle w_{1}^{{}}\text{ }\!\!|\!\!\text{ }{{w}_{1}}\rangle \otimes \text{ }\!\!|\!\!\text{ }{{U}_{q2}}{{w}_{2}}\rangle \overset{\langle w_{1}^{{}}\text{ }\!\!|\!\!\text{ }{{w}_{1}}\rangle \text{=}1}{\mathop{\text{?=}}}\,\langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }{{U}_{q2}}\text{ }\!\!|\!\!\text{ }{{w}_{2}}\rangle \\ & \\ & \langle w_{2}^{{}}U_{q2}^{\dagger }w_{1}^{{}}|{{U}_{k1}}{{w}_{1}}{{w}_{2}}\rangle \text{?=}\langle w_{2}^{{}}w_{1}^{{}}|{{U}_{q2}}{{U}_{k1}}|{{w}_{1}}{{w}_{2}}\rangle \\ & \to \langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }\otimes \langle w_{1}^{{}}\text{ }\!\!|\!\!\text{ }U_{q2}^{{}}{{U}_{k1}}|{{w}_{1}}\rangle \otimes \text{ }\!\!|\!\!\text{ }{{w}_{2}}\rangle \\ \end{align}

3.$$(|{{w}_{1}},{{U}_{2}}{{w}_{3}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}},{{U}_{2}}{{w}_{3}}|=\langle {{w}_{1}},{{w}_{3}}|U_{2}^{\dagger }$$

The above two formulas are my own. The confusion of the first formula is when the quantum operator is moved and it is found that the Bar and Ket of W1 are closely adjacent, so whether it can be eliminated by the inner product. The second formula wants to separate the operators in the formula, and I don't know whether the result is correct. The above rules are all inspired by a book "QUANTUM COMPUTING From Linear Algebra to Physical Realizations".

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2 Answers 2

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I believe your first equation to be correct, but I would alter the notation slightly to clarify. $$\langle w_1,w_2|w_1,U_2w_2\rangle$$ $$=\langle w_1,w_2|I\otimes U_2|w_1,w_2\rangle$$ $$=\langle w_1|I|w_1\rangle \otimes \langle w_2|U_2|w_2\rangle $$ $$=\langle w_2|U_2|w_2\rangle$$ with $I$ as the identiy matrix and, as you stated, $\langle w_1|I|w_1\rangle=\langle w_1|w_1\rangle= 1$

As for the second equation $$\langle w_1,U_2^\dagger w_2|U_1 w_1,w_2\rangle$$ $$=\langle w_1,w_2|U_1\otimes U_2|w_1,w_2\rangle$$ $$=\langle w_1|U_1|w_1\rangle \otimes \langle w_2|U_2|w_2\rangle$$

would make more sense to me, but I may be misinterpretting your q's and k's.

Note: It is convention to keep the systems in order. This is why I never wrote $\langle w_2,w_1|$, because system 1 is before system 2 whether in a Bra or a Ket.

Edit 1.0: For additional inquiries asked

Looking at the work that you did, you are really close! Your error happened when you calculated the tensor product of $\langle w_1,w_2|$ $$\langle w_1,w_2| = \langle w_1 \otimes w_2|$$ $$=(1/\sqrt{2} \;\; -1/\sqrt{2})\otimes (1/\sqrt{2} \;\; 1/\sqrt{2})$$ $$=(1/2 \;\;\; 1/2 \;\; -1/2\;\; -1/2)$$

Making this correction to your tensor product should allow you to calculate that $\langle w_1,w_2|w_1,U_2w_2\rangle =1$ as you got for the RHS of your equation.

Edit 2.0: Illustrating invisible tensor product

You have stated that $(w_1 w_2)^\dagger = w_2^\dagger w_1^\dagger$ in linear algebra, but that you believe this to be different than what has been done above, however if we insert the invisibe tensor product we will see that these are the same even in quantum Dirac notation. $$(w_1 w_2)^\dagger = w_2^\dagger w_1^\dagger$$ $$(w_1 \otimes w_2)^\dagger = w_2^\dagger \otimes w_1^\dagger$$ So using your $w_1 = (1/\sqrt{2} \:\:-1/\sqrt{2})^T$ and $w_2 = (1/\sqrt{2} \:\: 1/\sqrt{2})^T$ we get $$\begin{pmatrix} 1/2\\ 1/2 \\-1/2\\ -1/2\end{pmatrix}^\dagger = (1/\sqrt{2} \:\: -1/\sqrt{2}) \otimes (1/\sqrt{2} \:\: 1/\sqrt{2})$$ $$\begin{pmatrix} 1/2& 1/2 &-1/2 &-1/2\end{pmatrix}=\begin{pmatrix} 1/2& 1/2 &-1/2 &-1/2\end{pmatrix}$$

For further information on the tensor product, $\otimes$, look here

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  • $\begingroup$ I do not think so, can you check my following ans.? $\endgroup$ Aug 23, 2022 at 23:14
  • $\begingroup$ Thanks a lot for your reply. I am very curious about this magical movement rule of quantum systems. There seems to be a certain pattern. But I can't get it exactly from some basic textbooks because I'm not a quantum physics major. I'm trying to understand just from a linear algebra perspective, but a lot of times I don't understand it completely. Can you summarize the rules you deduced above? $\endgroup$ Aug 23, 2022 at 23:54
  • $\begingroup$ I've tried to answer your additional question in my edit. Hope this clear it up! $\endgroup$
    – PGibbon
    Aug 24, 2022 at 13:56
  • $\begingroup$ Thanks for your answer to the question. So $$(|{{w}_{1}},{{U}_{2}}{{w}_{3}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}},{{U}_{2}}{{w}_{3}}|=\langle {{w}_{1}},{{w}_{3}}|U_{2}^{\dagger }$$This is indeed different from the conjugate transpose of a product in linear algebra: $${{({{w}_{1}}{{w}_{3}})}^{\dagger }}=w_{3}^{\dagger }w_{1}^{\dagger }$$ $\endgroup$ Aug 24, 2022 at 14:56
  • $\begingroup$ For those seeking further information this conversation got continued on another post, quantumcomputing.stackexchange.com/questions/27921/… $\endgroup$
    – PGibbon
    Aug 29, 2022 at 12:43
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The issue is \begin{align} & (|{{w}_{1}}...{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}}...{{w}_{n}}|? \\ & (|{{w}_{1}}...{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{n}}...{{w}_{1}}|? \\ & (|{{w}_{1}}{{U}_{1}}{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{1}}U_{1}^{\dagger }{{w}_{n}}|? \\ & (|{{w}_{1}}{{U}_{1}}{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{n}}U_{1}^{\dagger }{{w}_{1}}|? \\ \end{align}

if we chose $(|{{w}_{1}}....{{w}_{n}}\rangle {{)}^{\dagger }}=\langle {{w}_{n}}...{{w}_{1}}|$ \begin{align} & \langle w_{2}^{{}}w_{1}^{{}}\text{ }\!\!|\!\!\text{ }{{w}_{1}}{{U}_{q2}}{{w}_{2}}\rangle \text{=?}\langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }{{U}_{q2}}\text{ }\!\!|\!\!\text{ }{{w}_{2}}\rangle \\ & |{{w}_{2}}\rangle =\left( \begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{matrix} \right),|{{w}_{1}}\rangle =\left( \begin{matrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ \end{matrix} \right),{{U}_{q2}}=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right) \\ & \langle {{w}_{2}}{{w}_{1}}|=\left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{matrix} \right)\otimes \left( \begin{matrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{matrix} \right)=\left( \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \end{matrix} \right) \\ & \text{ }\!\!|\!\!\text{ }{{w}_{1}}{{U}_{q2}}{{w}_{2}}\rangle =\left( \begin{matrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ \end{matrix} \right)\otimes \left( \begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{matrix} \right)=\left( \begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \\ \end{matrix} \right) \\ & \langle w_{2}^{{}}w_{1}^{{}}\text{ }\!\!|\!\!\text{ }{{w}_{1}}{{U}_{q2}}{{w}_{2}}\rangle =\left( \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \end{matrix} \right)\left( \begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \\ \end{matrix} \right)=0 \\ & \langle w_{2}^{{}}\text{ }\!\!|\!\!\text{ }{{U}_{q2}}\text{ }\!\!|\!\!\text{ }{{w}_{2}}\rangle =\left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{matrix} \right)\left( \begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{matrix} \right)=1 \\ \end{align}

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  • $\begingroup$ If you could add this as an edit to your initial question it would make more sense than having it here as a potential answer. Thanks! @R-X Zhao $\endgroup$
    – PGibbon
    Aug 24, 2022 at 13:55

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