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Leonard
Leonard
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Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \text{and} \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle. $$$$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \& \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle, $$ Here,where non-trivial means that $ (a_{1},b_{1}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $ and $ (a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $$ (a_{1},b_{1}),(a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.

Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \text{and} \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle. $$ Here, non-trivial means that $ (a_{1},b_{1}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $ and $ (a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.

Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \& \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle, $$ where non-trivial means that $ (a_{1},b_{1}),(a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.

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Leonard
Leonard
  • 816
  • 5
  • 15

Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \text{and} \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle. $$ Here, non-trivial means that $ (a_{1},b_{1}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $ and $ (a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.