Phase? Phase difference? What is the mathematical logic of global phase and local phase?

In quantum mechanics, both global phase and local phase are properties of a particular quantum state. For the quantum state of a single qubit, we usually express it as $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ , where$\alpha$和$\beta$ is complex and satisfies$|\alpha {|}^2+|\beta {|}^2=1$ to ensure the uniformity of the quantum state.

Global Phase: The global phase of a quantum state is relative to a specific reference state (usually ∣ $|0⟩$ ) defined. If we put$|\psi ⟩$ multiplied by a complex phase factor$e^{i\varphi }$ , the obtained quantum state$e^{i\varphi }|\psi ⟩$is and the original quantum state$|\psi ⟩$ describes the same physical state, but the global phase is different. The global phase is not measurable in physical observations.

Local Phase: The local phase of a quantum state is relative to its own ground state (e.g. $|0⟩$ and$|1⟩$ ) defined. For the quantum state$|\psi ⟩=\alpha |0⟩+\beta |1⟩$ , the local phase usually refers to$\alpha$和β \betaThe phase difference of$\beta$$arg\left(b\right)-\arg \left(\alpha \right)$ .$\_$The local phase can in some cases be measured by quantum interference experiments.

Calculated as follows:

Global phase: ${\varphi }_g=arg\left(a\right)$

Local phase: ${\varphi }_l=arg\left(b\right)-arg\left(a\right)$

Here $arg\left(x\right)$ is the plural$Argument\; function\; of\; x$ , returns$The\; phase\; of\; x$ (in$-p$ to$within\; the\; range\; of\; \pi$ ).

Note that these are theoretical definitions and calculation methods. In an actual quantum system, since the global phase is unmeasurable in physical observations, we usually only focus on the local phase. At the same time, the local phase is not directly measurable, but needs to be measured indirectly through techniques such as quantum interference.

$i=\arg (z)=\arctan \left(\frac{y}{x}\right)$

This expression is to calculate two quantum states $|{\psi }_1⟩$和$|{\psi }_2⟩$ The phase between. Molecule$⟨p_1|{\psi }_2⟩$ is the inner product between two states, and the denominator$|⟨p_1|{\psi }_2⟩|$ is the absolute value of the inner product.

内立$⟨p_1|{\psi }_2⟩$ is a complex number whose absolute value gives the overlap (or similarity) of two states, and whose phase gives the phase difference of the two states. Thus, the result of this expression is a complex number modulo 1 whose phase is equal to the phase difference of the two states.

This expression is often used in quantum mechanics to calculate the phase relationship between two states. For example, if we have two states $|{\psi }_1⟩=|0⟩$和$|{\psi }_2⟩=e^{i\varphi }|0⟩$, then⟨$\frac{⟨p_1|{\psi }_2⟩}{|⟨p_1|{\psi }_2⟩|}=e^{i\varphi }$ gives$|{\psi }_2⟩relative$ to$|{\psi }_1⟩$ phase.

In this particular example, the global phase has no clear meaning, because we don't have a fixed reference state to define the global phase. The global phase usually refers to a quantum state relative to a fixed reference state (such as $|0⟩$ or$|+⟩$ ).

However, we can calculate the relative phase between two states, that is, $|{\psi }_1⟩=\frac{1}{\sqrt{2}}(|0⟩+|1⟩)$和$|{\psi }_2⟩=\frac{1}{\sqrt{2}}(|0⟩+e^{i\varphi }|1⟩)$phase difference between. This phase difference is obtained by comparing$|{\psi }_1⟩$和$|{\psi }_2⟩$中$|1⟩$ coefficient to get.

在$|{\psi }_1⟩$中,$The\; coefficient\; of\; |1⟩$ is$\frac{1}{\sqrt{2}}$; while in $|{\psi }_2⟩$中,$The\; coefficient\; of\; |1⟩$ is$\frac{e^{i\varphi }}{\sqrt{2}}$. Therefore, the relative phase between two states is $e^{The\; phase\; of\; i\varphi }$ , that is,$\varphi$ .

So, relative to $|{\psi }_1⟩$，$|{\psi }_{2}The\; relative\; phase\; of\; ⟩$ is$\varphi$ .
Suppose we have a qubit whose state is$|\psi ⟩=\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩$ .$\_$This qubit is in a superposition state, we measure it in the ground state { ∣ 0 ⟩ |0\rangle $|0⟩$ ,$|1⟩$ } state, get$|0⟩$ and$The\; probability\; of\; |1⟩$ is$1/2$。

Now, we impose a global phase on this qubit, resulting in a new state $|{\psi }^{\prime }⟩=e^{i\theta }|\psi ⟩=e^{i\theta }(\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩)=\frac{1}{\sqrt{2}}{e}^{i\theta }|0⟩+\frac{1}{\sqrt{2}}{e}^{i\theta }|1⟩$。

You will see that although we change the phase of the quantum state, we measure the new quantum state $|{\psi }^{\prime }⟩$in the ground state { ∣ 0 ⟩ |0\rangle $|0⟩$ ,$|1⟩$ } state, get$|0⟩$ and$The\; probability\; of\; |1⟩$ is still$1/2$ . This means that the change of the global phase does not affect the physical observation results.
However, if we impose a local phase on this qubit, only changing$|1⟩$ phase, get a new state$|{\psi }^{\prime \prime }⟩=\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}{e}^{i\varphi }|1⟩$. Then, when we perform some specific measurements (such as measuring the probability distribution on other bases, or doing a quantum interference experiment), we can see the influence of this phase change.

Let's look at a concrete example. Suppose we have a qubit whose initial state is $|\psi ⟩=\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩$ .$\_$We can verify that this qubit is in the ground state$|+⟩$ and$|-⟩$ under the measurement probability is 50%.

Now, we impose a local phase on this qubit, changing only $|1⟩$ phase, get a new state$|{\psi }^{\prime }⟩=\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}{e}^{i\varphi }|1⟩$。

We can calculate that this new quantum state is in the ground state $|+⟩$ and$|-⟩$ under the measurement probability. In order to simplify the calculation, we assume$\varphi =\pi$ , then the new quantum state becomes$|{\psi }^{\prime }⟩=\frac{1}{\sqrt{2}}|0⟩-\frac{1}{\sqrt{2}}|1⟩$。

In this case, we can find that the new quantum state is in the ground state $|+⟩$ The measurement probability becomes 0%, while in the ground state$|-⟩$ The measurement probability becomes 100%. This shows that the change of the local phase affects the measurement results of the quantum state in the new basis.

You can try to calculate ϕ \phi for other values$\varphi$ , it will be found that the change of the local phase will indeed affect the measurement results of the quantum state in the new basis.

Initial state $|\psi ⟩=\frac{1}{\sqrt{2}}|0⟩+\frac{1}{\sqrt{2}}|1⟩$ is actually$|+⟩$ state. So in$|+⟩$ and$|-⟩$ is measured under the basis to get$|+⟩$ with probability 100%, and get$|-⟩$ has a probability of 0%.

When we pair $|1⟩$ apply a$After\; the\; local\; phase\; of\; \pi$ , the new state becomes$|{\psi }^{\prime }⟩=\frac{1}{\sqrt{2}}|0⟩-\frac{1}{\sqrt{2}}|1⟩$ , which is actually$|-⟩$ state. So in$|+⟩$ and$|-⟩$ is measured under the basis to get$|+⟩$ has a probability of 0%, and one gets$|-⟩$ has a probability of 100%.

In this example, we have a two-qubit state $|\psi ⟩=\sqrt{\frac{1}{2}}(|01⟩-|10⟩)$ . We can see that this state is already a normalized state, because$|01⟩$ and$The\; sum\; of\; the\; squares\; of\; the\; magnitudes\; of\; the\; probability\; amplitudes\; of\; |10⟩$ is 1.

A global phase is a phase that acts on the entire quantum state. For example, we can add a global phase by multiplying by a complex number, resulting in a new quantum state $|{\psi }^{\prime }⟩=e^{i\theta }|\psi ⟩=\sqrt{\frac{1}{2}}{e}^{i\theta }(|01⟩-|10⟩)=\sqrt{\frac{1}{2}}(e^{i\theta }|01⟩-e^{i\theta }|10⟩)$. This is a global phase because it applies to the entire quantum state.

A local phase is a phase that only acts on some quantum states. For example, we can just $|01⟩$ add a phase to get a new quantum state$|{\psi }^{\prime \prime }⟩=\sqrt{\frac{1}{2}}(e^{i\varphi }|01⟩-|10⟩)$ . This is a local phase because it only works on$|01⟩$ , but not on$|10⟩$ on.

Changes in the global phase do not affect physical observations, whereas changes in local phases may, for example in quantum interference.

Suppose an initial state of a single qubit $|\psi ⟩=|0⟩$ , after the following operations, the measurement results are obtained∣$|1⟩$：

将$|\psi ⟩$ Apply a Hadamard gate:$H|\psi ⟩=\frac{1}{\sqrt{2}}(|0⟩+|1⟩)$。
将$|\psi ⟩$ is used as a control bit, and a target bit$|-⟩=\frac{1}{\sqrt{2}}(|0⟩-|1⟩)$ as the target bit, perform CNOT operation:$CNO{T}_{|\psi ⟩,|-⟩}\frac{1}{\sqrt{2}}(|0⟩|-⟩+|1⟩|+⟩)$ .
Measure the first qubit. If the measurement result is$|0⟩$ , then the target bit measurement result is$|-⟩$ ; If the measurement result is$|1⟩$ , then the target bit measurement result is$|+⟩$。
$$CNO{T}_{|\psi ⟩,|-⟩}\frac{1}{\sqrt{2}}(|0⟩|-⟩+|1⟩|+⟩)=\frac{1}{\sqrt{2}}(|0⟩|-⟩-|1⟩|+⟩)$$

In this example, phase inversion occurs in step 2 when the control bit $The\; state\; of\; |\psi ⟩$ is$|1⟩$ , the phase of the target bit will be reversed.