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A lot of books and papers in QM use $\frac{1}{\sqrt2}$ in equations. If we want to calculate the intensity of a light or some probability, I see this irrational value every where. why do we use this value? I don't like getting philosophical in physics but I have been asking this question for a while. Is it because of normalization, such that the sum of some squares of two vectors have a value equal to 1? Are we considering vectors with an angle of $\frac{\pi}{4}$? How does this concept relate to probability?

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  • $\begingroup$ @Noone I believe philosophy would take the math out of physics. I like to think about particles, and how they behave but to ask why they act like that will lead to a non mathematical destination. For me, the fun is in "how?" and not "why?" $\endgroup$
    – Perfectoid
    Jun 22 at 19:17

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The common use of $\frac{1}{\sqrt{2}}$ nothing more complicated than that it's one of the easiest examples to use.

If I flip a fair coin, the output probabilities are 50/50 for heads and tails.

What is the most similar thing to a fair coin for a quantum system that is in a superposition? It would be a state with 2 possible outcomes where the probability is 50/50 for each measured outcome.

States are represented by their probability amplitudes, and the square of these probability amplitudes are the probability, then the state could have the form:

$$|\psi \rangle = \frac{1}{\sqrt{2}}\left(|H \rangle + |T \rangle\right)$$

similarly this state also has the same probability outcomes:

$$|\psi \rangle = \frac{1}{\sqrt{2}}\left(|H \rangle - |T \rangle\right)$$

and in fact any complex phase can have the same probabilities, which can be written generally as:

$$|\psi \rangle = \frac{1}{\sqrt{2}}\left(|H \rangle + e^{i\phi}|T \rangle\right)$$

As you can see, unlike the regular coin, there is an infinite set of possible states that have probability amplitudes that produce a 50/50 fair-coinflip probability. So the sqrt of two is in there simply because its the squareroot of $\frac{1}{2}$. It's nothing more complicated than that it's one of the easiest examples to use.

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The superposition of two eigenstates $|0\rangle$ is of the form $\alpha|0\rangle+\beta|1\rangle$ with $\alpha,\,\beta$ complex numbers whose square moduli sum to $1$. A measurement obtains $|0\rangle$ with probability $|\alpha|^2$ or $|1\rangle$ with probability $|\beta|^2$. In the special case where these outcomes are equally likely and have a relative phase of $0$ ($\pi$) to give a symmetric (antisymmetric) state, the superposition is of the form $e^{i\theta}\frac{|0\rangle\pm|1\rangle}{\sqrt{2}}$ ($\pm$ is $+$ in phase, $-$ in antiphase), where $\theta$ is real and without loss of generality $0$.

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  • $\begingroup$ Thank you for the explanation. $\endgroup$
    – Perfectoid
    Jun 22 at 19:19
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Quantum mechanics computes probability amplitudes. To get a probability, you take the norm squared of a probability amplitude.

Two outcomes which are equally likely will each have probability $\frac{1}{2}$. This means each outcome will have probability amplitude $\frac{1}{\sqrt{2}}$. Actually, they could have probability amplitude $\frac{e^{i\theta}}{\sqrt{2}}$ for any real $\theta$.

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  • $\begingroup$ Thank you! That's something basic but very clarifying. $\endgroup$
    – Perfectoid
    Jun 23 at 18:18

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