Suppose I have the state $|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ that I want to measure in an arbitrary basis $$|A\rangle = \alpha|0\rangle + \beta|1\rangle \text{ and } |B\rangle = \beta^*|0\rangle - \alpha^*|1\rangle$$

From my understanding, if I measure $|\psi\rangle$, the probability of seeing $|A\rangle$ is $$\langle A | \psi\rangle^2$$ But when I try to compute $\langle A | \psi\rangle$, I get

\begin{align*} \langle A | \psi\rangle &= \frac{1}{\sqrt{2}}(\alpha^*\langle 0 | + \beta^* \langle 1 |)(|01\rangle - |10\rangle)\\ &= \frac{1}{\sqrt{2}}(\alpha^*\langle 0 | + \beta^* \langle 1 |)(|01\rangle - |10\rangle) \\ &= \frac{1}{\sqrt{2}}(\alpha^*\langle 0 | + \beta^* \langle 1 |)(|0\rangle\otimes|1\rangle - |1\rangle\otimes|0\rangle) \end{align*}

I assume I can distribute, so I get terms like $\alpha^*\langle 0 |\big(|0\rangle\otimes|1\rangle\big)$.

But how does one take an inner product between $|0\rangle$ and $|0\rangle\otimes|1\rangle$, when the latter of the two is an element of a tensor product space of different dimension as $|0\rangle$?

  • $\begingroup$ You are right. You cannot take the inner product between $|0\rangle$ and $|0\rangle \otimes |1\rangle$ as, by definition, the inner product is: $$\langle \rangle : \mathbb{V}^2 \mapsto \mathbb{C}$$ The vector $|0\rangle \otimes |1\rangle$ belongs in the tensor product $\mathbb{V} \otimes \mathbb{V}$, not $\mathbb{V}$. $\endgroup$ – Andreas Mastronikolis Oct 5 at 16:31
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    $\begingroup$ @AndreasMastronikolis That's what I thought. In that case, how do we calculate measurement results and measurement probabilities? $\endgroup$ – Tiwa Aina Oct 5 at 16:32
  • $\begingroup$ To be honest, I am not currently sure as I am not completely acquainted with product spaces (yet). But here is how I am thinking: You are right in your assertion that the probability of catching $|\psi \rangle$ in the basis state $|A \rangle$ is: $$| \langle A| \psi \rangle|^2$$ The only thing that is left is to find a vector in $\mathbb{V} \otimes \mathbb{V}$ that will 'represent' (I know I am speaking loosely here) $|A\rangle$. My best guess would be: $$|A\rangle \otimes | A \rangle$$ but I am not sure. $\endgroup$ – Andreas Mastronikolis Oct 5 at 16:43
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    $\begingroup$ Bare in mind, that the quantity $$\left\lvert \left( \langle A| \otimes \langle A| \right) | \psi \rangle \right\rvert^2$$ is the probability of catching particle 1 in $|A\rangle$ and particle 2 in $|A\rangle$. $\endgroup$ – Andreas Mastronikolis Oct 5 at 16:53

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