spins_unit_operators_and_measurements.ppt Page 14

• The completeness relation basically tells us that any vector can be expanded in terms of an orthonormal basis.
• In terms of spin-$\frac{1}{2}$ systems, the completeness relations can be written in the z-basis as

$$\vert +\rangle \langle + \vert \; + \; \vert- \rangle\langle -\vert=1 \; \; . $$

Be sure to note that this can be done in any basis, but you have to pick one and stick to it to satisfy the completeness relation.

• To prove this, we can take the expression on the left of the equal sign and “probe” it to find more information by placing an arbitrary bra $\langle\psi \vert$ and a ket $\vert\psi \rangle$ on the left and right side, respectively. This looks like

$$\langle\psi \vert\Big(\: \vert +\rangle\langle+ \vert \; \; + \; \; \vert- \rangle\langle -\vert \: \Big)\vert\psi \rangle \; \; .$$

Now, we can FOIL the expression to expand, which yields

$$\langle\psi \vert +\rangle\langle+ \vert\psi \rangle \; \; + \; \; \langle\psi \vert- \rangle\langle -\vert\psi \rangle \; \; . $$

• Notice that in each term, we have some bra-ket combination times its complex conjugate. So, we can re-write each term as

$$\left\vert \langle+ \vert\psi \rangle\right\vert ^{2} \; + \; \left\vert\langle -\vert\psi \rangle\right\vert^{2} \; \; . $$

Now, if the term on the left is the probability of finding $\vert \psi \rangle$ in the $\vert + \rangle$ state, and the right term is the probability of finding $\vert \psi \rangle$ in the $\vert- \rangle$ state, then the total probability must add up to 1. So, we have

$$\left\vert\langle+ \vert\psi \rangle\right\vert^{2} \; + \; \left\vert\langle -\vert\psi \rangle\right\vert^{2}= 1 \; \; . $$

• If we look at the expression that contains our $\psi$ probes and compare it to the final expression we have, what we have found is that the probability of finding $\Big(\: \vert +\rangle\langle+ \vert \; + \; \vert- \rangle\langle -\vert \: \Big)\vert\psi \rangle$ in the $\vert\psi \rangle$ state is equal to one. The only way this will hold true is if

$$\Big(\: \vert +\rangle\langle+ \vert \; + \; \vert- \rangle\langle -\vert\: \Big)\vert\psi \rangle= \vert\psi \rangle \; \; .$$

The only way this will hold true is if

$$\vert +\rangle\langle+ \vert \; + \; \vert- \rangle\langle -\vert=I \; \; .$$