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Partial measurement rule

Quantum teleportation

So far, measurement of a quantum state $\ket{\psi} = \sum_x \alpha_x \ket{x}$ meant obtaining the classical state $\ket{x}$ with probability $|\alpha_x|^2$.

This is what we call a **standard basis measurement** or a **computational basis measurement**: the outcomes are the standard basis (a.k.a. computational basis) vectors
$$
\begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \qquad \cdots \qquad \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix}
$$

But often we'd like to measure a quantum state in a different basis, where the outcomes are quantum states that are *not* computational basis states.

Another basis for $\C^2$ is $\{ \ket{+}, \ket{-} \}$ (called the **diagonal basis**). This is an orthonormal basis, and that means we can also measure with respect to this basis.

Rewrite $\ket{\psi}$ as a linear combination of the $\{ \ket{+}, \ket{-} \}$ basis: $$\ket{\psi} = \alpha \frac{\sqrt{2}}{2} \Big( \ket{+} + \ket{-} \Big) + \beta \frac{\sqrt{2}}{2} \Big ( \ket{+} - \ket{-} \Big) $$

The probability of getting $\ket{+}$ outcome is thus $$ \Big( \frac{\sqrt{2}}{2} \Big)^2 \cdot \Big | \alpha + \beta \Big|^2 $$ and similarly the probability of $\ket{-}$ outcome is $$ \Big( \frac{\sqrt{2}}{2} \Big)^2 \cdot \Big | \alpha - \beta \Big|^2 $$

There's also a geometric way to think about it: $$\Pr \Big [ \text{measuring $\ket{\psi}$ in diagonal basis yields $\ket{+}$} \Big] = \text{overlap of $\ket{\psi}$ and $\ket{+}$}$$

There's also a geometric way to think about it: $$\Pr \Big [ \text{measuring $\ket{\psi}$ in diagonal basis yields $\ket{+}$} \Big] = \Big | \Big \langle + \mid \psi \Big \rangle \Big|^2$$ Similarly, $$\Pr \Big [ \text{measuring $\ket{\psi}$ in diagonal basis yields $\ket{-}$} \Big] = \Big | \Big \langle - \mid \psi \Big \rangle \Big|^2$$

Let $\ket{\psi} \in \C^d$ be a quantum state. Let $B = \{ \ket{b_1},\ldots,\ket{b_d} \}$ be an **orthonormal basis** for $\C^d$. Then measuring $\ket{\psi}$ with respect to basis $B$ yields outcome $\ket{b_j}$ with probability
$$
\Big | \Big \langle b_j \Big | \, \psi \Big \rangle \Big |^2
$$
and the post-measurement state is $\ket{b_j}$.

How do you actually measure in a different basis? For example in Qiskit you only get the ability to measure in the standard basis.

Measuring $\ket{\psi}$ in a basis $B$ is ** equivalent** to

- Apply a unitary $V$ that maps $B$ to standard basis (i.e. $\ket{b_j} \to \ket{j}$).
- Measure in standard basis.

Let's combine the two concepts! Let $\ket{\psi} = \sum_{i,j} \alpha_{ij} \ket{i} \otimes \ket{j}$ denote a two-qubit state. Say we measure the first qubit with respect to basis $\{ \ket{b_0}, \ket{b_1} \}$.

**Measurement rule**: Probability of obtaining outcome $\ket{b_0}$ is the ** length squared** of the vector
G
$$
\ket{a} = (\bra{b_0} \otimes I)\ket{\psi} = \sum_{i,j} \alpha_{ij} \langle b_0 \mid i \rangle \, \, \ket{j}.
$$
Note this is a vector in $\C^2$, not $\C^2 \otimes \C^2$. It's a

The probability is thus:
$$
\Big \| \, \ket{a} \, \Big \|^2 = \sum_{j} \Big | \sum_i \alpha_{ij} \langle b_0 \mid i \rangle \Big |^2
$$
The **post-measurement** state of both qubits is
$$
\ket{b_0} \otimes \frac{ \ket{a}}{ \| \ket{a} \| }.
$$

Let's work through an example:
$$
\ket{\psi} = \ket{\psi_1} \otimes \ket{\psi_2}.
$$
Measure the *first* qubit in the *diagonal basis*.

- What is the probability the outcome is $\ket{+}$? Or $\ket{-}$?
- What is the post-measurement state in either case?

Let's work through an example:
$$
\ket{\psi} = \sqrt{\frac{2}{3}} \ket{01} - \sqrt{\frac{1}{3}} \ket{10}.
$$
Measure the *first* qubit in the *diagonal basis*.

- What is the probability the outcome is $\ket{+}$? Or $\ket{-}$?
- What is the post-measurement state in either case?

Popular science version: can't exactly know both the position and momentum of a particle simultaneously.

In quantum information theory terms: it is not possible for a qubit $\ket{\psi} \in \C^2$ to be simultaneously determined in both the *standard* basis and the *diagonal basis*.

In other words, if measuring $\ket{\psi}$ in standard basis yields a deterministic outcome, then it *cannot* have a deterministic outcome if measured according to diagonal basis.

* Important point*: it's not about what happens if you

* Important point*: it's not about what happens if you

It's reasoning about **counterfactual scenarios**: measuring $\ket{\psi}$ in the standard basis, **or** measuring $\ket{\psi}$ in the diagonal basis.

We say that the standard basis and diagonal basis are *incompatible* or *complementary*.

In quantum physics, the position and momentum of a particle correspond to incompatible measurements!

There's also a quantitative version that you may have to prove in the homework, meaning that there is a tradeoff between the uncertainty in the standard basis versus the uncertainty in the diagonal basis.

In 1935, Einstein, Podolsky, and Rosen published a paper called

Can Quantum-Mechanical Description of Physical

Reality be Considered Complete?

The EPR thesis:

Quantum mechanics may be very good at predicting outcomes of experiments, but it cannot be a

completedescription of Nature.

The reason they thought this was because of a thought experiment.

Alice and Bob are in far-away galaxies and share the EPR state $$ \ket{\Phi} = \frac{1}{\sqrt{2}} \Big( \ket{00} + \ket{11} \Big). $$ Consider two possible experiments:

**Experiment A**: Alice measures her qubit in the standard basis $\{ \ket{0} ,\ket{1} \}$**Experiment B**: Alice measures her qubit in the diagonal basis $\{ \ket{+} ,\ket{-} \}$

Alice gets outcome

- $\ket{0}$ with probability $1/2$, and the post-measurement state is $\ket{00}$.
- $\ket{1}$ with probability $1/2$, and the post-measurement state is $\ket{11}$.

To calculate the probability of getting outcome $\ket{+}$, we use the partial measurement + nonstandard basis rules: first, compute the vector

$$ \ket{v_+} = \Big (\bra{+} \otimes I\Big )\, \ket{\Phi} = \frac{1}{\sqrt{2}} \Big( \langle + \mid 0 \rangle \, \ket{0} \,\, + \,\, \langle + \mid 1 \rangle \, \ket{1} \Big) $$$$ = \frac{1}{\sqrt{2}} \Big( \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1} \Big) = \frac{1}{2} \ket{0} + \frac{1}{2} \ket{1}. $$The squared length is $ \| \ket{v_+} \|^2 = \frac{1}{2^2} + \frac{1}{2^2} = \frac{1}{2}$.

The post-measurement state is then $\ket{+} \otimes \ket{v_+}/\sqrt{2} = \ket{+} \otimes \ket{+}$.

Alice and Bob are in far-away galaxies and share the EPR state $\ket{\Phi} = \frac{1}{\sqrt{2}} \Big( \ket{00} + \ket{11} \Big)$. If Alice measures in standard basis, after seeing her result she knows exactly what state Bob's qubit is in -- even if Bob's qubit is zillions of lightyears away.

- If Alice did Experiment A and got outcome (say) $\ket{0}$, then it must have been Bob's qubit was really $\ket{0}$ all along.

EPR's conclusion: Quantum Mechanics can make the right statistical predictions. But it's just a ** mathematical model** that doesn't actually describe the way Reality works.

There is a ** deeper** classical theory -- called

- Reproduces the same statistics as Quantum Mechanics
- But has hidden variables that describes the intrinsic state of particles.
- Respects the speed of light limit.

What do you think of Einstein's reasoning? It is a paradox?

John Bell's answer and a game with entanglement.