$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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overview
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In quantum, we require states to be normalized. This means that adding up probability for a state to be measured as it's allowed values, it has to add up to one.
Unnormalized states need to be multiplied by a normalization constant.

Significance

For instance, for measuring spin, let $a$ denote the probabilty of measuring spin up, and $b$ denote the probability of measuring spin down. Since the particle must be in the state spin up or spin down, that means we require $a+b=1$.
For something continuous such as position, if we know a particle must be in a certain area, then the integral of finding the probability over that adds up to $1$.

Derivation

Let $\ket{\psi}$ be a state such that its possible outcomes are discrete states $\ket{\phi_1}, \dots, \ket{\phi_n}$. We require $$\begin{align*} \sum_{k=1}^{n}\braket{\phi_k|\psi} =1 \end{align*}$$
Let $\ket{\psi}$ be a state such that its possible measurement outcomes are continuous. Then we require

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Associated Concepts

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In quantum, we require states to be normalized. This means that adding up probability for a state to be measured as it's allowed values, it has to add up to one.
Unnormalized states need to be multiplied by a normalization constant.

Significance

For instance, for measuring spin, let $a$ denote the probabilty of measuring spin up, and $b$ denote the probability of measuring spin down. Since the particle must be in the state spin up or spin down, that means we require $a+b=1$.
For something continuous such as position, if we know a particle must be in a certain area, then the integral of finding the probability over that adds up to $1$.

Derivation

Let $\ket{\psi}$ be a state such that its possible outcomes are discrete states $\ket{\phi_1}, \dots, \ket{\phi_n}$. We require $$\begin{align*} \sum_{k=1}^{n}\braket{\phi_k|\psi} =1 \end{align*}$$
Let $\ket{\psi}$ be a state such that its possible measurement outcomes are continuous. Then we require

Related Experiments

Coming soon

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts
FullPage
overview
significance
derivation
related experiments
associated concepts
In quantum, we require states to be normalized. This means that adding up probability for a state to be measured as it's allowed values, it has to add up to one.
Unnormalized states need to be multiplied by a normalization constant.

Significance

For instance, for measuring spin, let $a$ denote the probabilty of measuring spin up, and $b$ denote the probability of measuring spin down. Since the particle must be in the state spin up or spin down, that means we require $a+b=1$.
For something continuous such as position, if we know a particle must be in a certain area, then the integral of finding the probability over that adds up to $1$.

Derivation

Let $\ket{\psi}$ be a state such that its possible outcomes are discrete states $\ket{\phi_1}, \dots, \ket{\phi_n}$. We require $$\begin{align*} \sum_{k=1}^{n}\braket{\phi_k|\psi} =1 \end{align*}$$
Let $\ket{\psi}$ be a state such that its possible measurement outcomes are continuous. Then we require

Related Experiments

Coming soon

Associated Concepts

Coming soon
In quantum, we require states to be normalized. This means that adding up probability for a state to be measured as it's allowed values, it has to add up to one.
Unnormalized states need to be multiplied by a normalization constant.

Significance

For instance, for measuring spin, let $a$ denote the probabilty of measuring spin up, and $b$ denote the probability of measuring spin down. Since the particle must be in the state spin up or spin down, that means we require $a+b=1$.
For something continuous such as position, if we know a particle must be in a certain area, then the integral of finding the probability over that adds up to $1$.

Derivation

Let $\ket{\psi}$ be a state such that its possible outcomes are discrete states $\ket{\phi_1}, \dots, \ket{\phi_n}$. We require $$\begin{align*} \sum_{k=1}^{n}\braket{\phi_k|\psi} =1 \end{align*}$$
Let $\ket{\psi}$ be a state such that its possible measurement outcomes are continuous. Then we require

Related Experiments

Coming soon

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts