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I would like to check if state

$$ |\psi\rangle = \cos{\theta}|0\rangle + i \sin{\theta}|1\rangle $$

is properly defined.

But when I calculated $\langle \psi | \psi \rangle$ is get $$ \cos^2{\theta} - \sin^2{\theta} $$

so I am facing sin and cos.

What is wrong in my calculation?

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  • $\begingroup$ I want to make inner product of each state with itself to show the value is 1(valid state) or not. @MartinVesely $\endgroup$
    – Ba. Taj
    Jan 15 '20 at 19:26
  • $\begingroup$ Yes right @MartinVesely $\endgroup$
    – Ba. Taj
    Jan 15 '20 at 20:24
  • $\begingroup$ I deleted my comments because now I understand. Please find answer below. $\endgroup$ Jan 15 '20 at 22:24
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Standard inner product on space $\mathbb{C}^{n}$ is defined as

$$ \langle a|b \rangle = \sum_{i=1}^{n} a_{i}b_{i}^{\dagger}, $$

where $b^{\dagger}$ is complex conjugate (i.e. for $x \in \mathbb{C}$: if $x = x_{re} + ix_{im}$ then $x^{\dagger} = x_{re} - ix_{im}$).

In your case $a_{1} = \cos(\theta)$ and $a_{2} = i\sin(\theta)$. Since you are interested in product $\langle a|a \rangle$, $a_{1}^{\dagger} = \cos(\theta)$ because it is a real number and $a_{2}^\dagger = -i\sin(\theta)$ you have

$$ \langle a|a \rangle = a_{1}a_{1}^{\dagger}+a_{2}a_{2}^{\dagger} = \cos^{2}(\theta) -i^{2}\sin^{2}(\theta) = \cos^2(\theta) + sin^{2}(\theta) = 1 $$

Alternatively you can use normalization condition $|a|^2+|b|^2 = 1$:

$$|a|^2 = |\cos(\theta)|^2 = \cos^{2}(\theta)$$

and

$$|b|^2 = |i\sin(\theta)|^2 = |i|^2|\sin^{2}(\theta)| = 1^2\cdot \sin^{2}(\theta)$$

And again you have $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$.

Overall, your quatum state is properly defined.

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You asked "What is wrong in my calculation?"

My guess is that your mistake is in forgetting that $\langle \psi | = \cos{\theta}\langle 0| - i \sin{\theta}\langle 1|$ instead of $\cos{\theta}\langle 0| + i \sin{\theta}\langle 1|$ (because $i$ becomes $-i$ when you take the complex conjugate).

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