Admin¶
- Pset1 due Friday, 11:59pm.
- Must submit PDF to Gradescope, Jupyter notebook to Courseworks.
- Everybody has a budget of 5 late days.
- Henry, please record the lecture!
Last Time¶
Partial measurement rule
Quantum teleportation
Nonstandard Measurements¶
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 $$ \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \qquad \cdots \qquad \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix} $$
Nonstandard Measurements¶
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.
Let $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$ denote a qubit state. What happens if we measure in the diagonal basis?
Measuring in Diagonal Basis¶
Rewrite $\ket{\psi}$ as a linear combination of the $\{ \ket{+}, \ket{-} \}$ basis: $$\ket{\psi} = \alpha \frac{\sqrt{2}}{2} \Big( \ket{+} + \ket{-} \Big) + \beta \frac{\sqrt{2}}{2} \Big ( \ket{+} - \ket{-} \Big) $$
Measuring in Diagonal Basis¶
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 $$ \Big( \frac{\sqrt{2}}{2} \Big)^2 \cdot \Big | \alpha - \beta \Big|^2 $$
Another Perspective¶
There's also a geometric way to think about it: $$\Pr \Big [ \text{measuring $\ket{\psi}$ in diagonal basis yields $\ket{+}$} \Big] = \text{overlap of $\ket{\psi}$ and $\ket{+}$}$$
Another Perspective¶
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, $$\Pr \Big [ \text{measuring $\ket{\psi}$ in diagonal basis yields $\ket{-}$} \Big] = \Big | \Big \langle - \mid \psi \Big \rangle \Big|^2$$
General formula for measuring in a basis¶
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 $\ket{b_j}$.
Implementing a measurement in a different basis¶
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.
Probability of getting $\ket{j}$ in this new process is the same as getting $\ket{b_j}$ when measuring in basis $B$.
Partial Measurements plus Nonstandard Measurements¶
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 partial inner product.
Partial Measurements plus Nonstandard Measurements¶
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 $$ \ket{b_0} \otimes \frac{ \ket{a}}{ \| \ket{a} \| }. $$
Partial Measurements plus Nonstandard Measurements¶
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?
Partial Measurements plus Nonstandard Measurements¶
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?
Heisenberg Uncertainty Principle¶
Popular science version: can't exactly know both the position and momentum of a particle simultaneously.
Heisenberg Uncertainty Principle¶
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.
Heisenberg Uncertainty Principle¶
Important point: it's not about what happens if you sequentially measure the state $\ket{\psi}$ (what happens then?).
Heisenberg Uncertainty Principle¶
Important point: it's not about what happens if you sequentially measure the state $\ket{\psi}$ (what happens then?).
It's reasoning about counterfactual scenarios: measuring $\ket{\psi}$ in the standard basis, or measuring $\ket{\psi}$ in the diagonal basis.
Heisenberg Uncertainty Principle¶
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!
Heisenberg Uncertainty Principle¶
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.
The EPR Paradox¶
In 1935, Einstein, Podolsky, and Rosen published a paper called
Can Quantum-Mechanical Description of Physical
Reality be Considered Complete?
The EPR Paradox¶
The EPR thesis:
Quantum mechanics may be very good at predicting outcomes of experiments, but it cannot be a complete description of Nature.
The reason they thought this was because of a thought experiment.
The EPR Paradox¶
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 $\{ \ket{+} ,\ket{-} \}$
Experiment A¶
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 $\ket{11}$.
Experiment B¶
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}$.
Experiment B¶
The post-measurement state is then $\ket{+} \otimes \ket{v_+}/\sqrt{2} = \ket{+} \otimes \ket{+}$.
Similarly, the probability of getting outcome $\ket{-}$ is $\frac{1}{2}$ and the post-measurement state is $\ket{-} \otimes \ket{-}$.
The EPR Paradox¶
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.
The EPR Paradox¶
- 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.
- On the other hand, if Alice did Experimeent B and got outcome (say) $\ket{-}$, then it must have been Bob's qubit was really $\ket{-}$ all along.
- But Alice's choice of measurement (standard or diagonal) couldn't have made an instantaneous difference in intrinsic the state of Bob's qubit, right?
- Therefore Bob's qubit must have answers prepared for both Experiments simultaneously -- violating Heisenberg's Uncertainty Principle!
The EPR Paradox¶
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 local hiden variable theory -- that
- Reproduces the same statistics as Quantum Mechanics
- But has hidden variables that describes the intrinsic state of particles.
- Respects the speed of light limit.
The EPR Paradox¶
What do you think of Einstein's reasoning? It is a paradox?
Next time¶
John Bell's answer and a game with entanglement.