# Quantum Expectation Calculation with Orthogonal Projection

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?

## 1 Answer

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:

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

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*}

• 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 $|0\rangle\langle0| + |1\rangle\langle1|$? Commented Apr 11, 2020 at 22:04
• @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 $P_{M_2} = |+\rangle\langle+| = \frac{1}{2}(|0\rangle\langle0| + |0\rangle\langle1| + |1\rangle\langle0| + |1\rangle\langle1|)$ Commented Apr 11, 2020 at 22:26