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Let $\mathcal{H} =\mathbb{C}^2, \mathcal{M}_1 = \mathbb{C}|0\rangle$ with $|\psi\rangle = \alpha |0\rangle + \beta|1\rangle$. Show $Pr(\mathcal{M_1}) = |\alpha|^2.$

We know that $\mathcal{M_1}$ is a subspace of the Hilbert space $\mathcal{H}$, and that $Pr(\mathcal{M_1}) = \langle \psi| Proj_\mathcal{M}|\psi\rangle.$

I see that when we first evaluate $Pr(\mathcal{M_1}) = \langle \psi| Proj_\mathcal{M}|\psi\rangle$ we obtain

$\langle \psi| Proj_\mathcal{M}|\psi\rangle = \langle\alpha|0\rangle + \beta|1\rangle|Proj_{\mathbb{C}^0}|\alpha|0\rangle + \beta|1\rangle\rangle = \langle \alpha|0\rangle | \alpha|0\rangle + \beta|1\rangle\rangle$.

At this point I don't see how we can show the desired result. Am I not applying the projection operator with the given notion properly?

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The definition of projectors can be found on page 70 from M. Nielsen and I. Chuang textbook: Suppose $W$ is a $k$-dimensional vector subspace of the $d$-dimensional vector space $V$. Using the Gram–Schmidt procedure it is possible to construct an orthonormal basis $|1\rangle$,...$|d\rangle$ for $V$ such that $|1\rangle$,...$|k\rangle$ is an orthonormal basis for $W$. By definition:

$$P = \sum_{i = 1}^k | i \rangle \langle i |$$

is a projector onto the subspace $W$. End of the quote from the textbook.

So, for $H = \mathbb{C}^2$ $|0\rangle$ and $|1\rangle$ are orthonormal basis vectors and the $|0\rangle \langle 0|$ will be a valid projector into $M_1$ subspace:

$$Proj_{M_1} = |0\rangle \langle 0|$$

Then:

\begin{equation} Pr\left(M_1\right) = \langle \psi |Proj_{M_1}| \psi \rangle = \big(\alpha^* \langle 0| + \beta^* \langle 1 |\big)\big(|0\rangle \langle 0|\big) \big(\alpha | 0 \rangle + \beta | 1 \rangle\big) \\ =\big(\alpha^* \langle 0| 0 \rangle + \beta^* \langle 1 | 0 \rangle \big) \big(\alpha \langle 0 | 0 \rangle + \beta \langle 0 | 1 \rangle\big) = \alpha^* \alpha = |\alpha|^2 \end{equation}

where we took into account:

\begin{align*} &\text{ortnormality:}\quad \;\; \langle 0| 0 \rangle = \langle 1| 1 \rangle = 1; \quad \;\;\;\; \langle 0| 1 \rangle = \langle 1| 0 \rangle = 0 \\ &\text{$|\psi\rangle$ and $\langle \psi |$:} \qquad |\psi\rangle = \alpha | 0 \rangle + \beta | 1 \rangle; \qquad \langle \psi | = \alpha^* \langle 0| + \beta^* \langle 1 | \end{align*}

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  • $\begingroup$ If I had a subspace $\mathcal{M_2} = \mathbb{C}(|0\rangle + |1\rangle)$ would that mean that a projection for that space would be $|0\rangle\langle0| + |1\rangle\langle1|$? $\endgroup$
    – john smith
    Apr 11 '20 at 22:04
  • $\begingroup$ @johnsmith How I understand, according to the definition, we should find a different set of orthonormal basis vectors that span the $H$: $|+\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}} (|0\rangle - |1\rangle)$, and note that with one of this orthonormal basis ($|+\rangle$) we can span $M_2$ subspace: the projector here would be $P_{M_2} = |+\rangle\langle+| = \frac{1}{2}(|0\rangle\langle0| + |0\rangle\langle1| + |1\rangle\langle0| + |1\rangle\langle1|)$ $\endgroup$ Apr 11 '20 at 22:26

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