Two-level quantum systems

Two-dimensional vector

In this book, we use Dirac's vector notation used in quantum physics. Consider an arbitrary two-dimensional vector ∣$\mathit{\unicode[Book Antiqua]{x76}}$> as shown in figure 1.1 in the ket vector format: $| \mathit{\unicode[Book Antiqua]{x76}}\rangle =\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right]$ where $\mathit{\unicode[Book Antiqua]{x76}}$ ₁ and $\mathit{\unicode[Book Antiqua]{x76}}$ ₂ are the x- and y-components, respectively. For a Euclidean space, the coefficients are real values.

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Using a set of unit vectors, $| {e}_{1}=\left[\begin{array}{c}1\\ 0\end{array}\right]$ and $| {e}_{2}\rangle =\left[\begin{array}{c}0\\ 1\end{array}\right]$, the vector ∣$\mathit{\unicode[Book Antiqua]{x76}}$> can be expressed as

$$\begin{eqnarray}&&| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}| e{}_{1}\rangle +{\mathit{\unicode[Book Antiqua]{x76}}}_{2}| {e}_{2}\rangle ,\,{\rm{or}}\,| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}\left[\begin{array}{c}1\\ 0\end{array}\right]+{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right].\end{eqnarray} \tag{ 1.1 }$$

Define the bra vector: $\langle u| =\left[{u}_{1}{u}_{2}\right]$, and the inner (scalar) product of two vectors, <u∣ and ∣$\mathit{\unicode[Book Antiqua]{x76}}$>, can be defined as

$$\begin{eqnarray}&&\langle u| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={u}_{1}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}+{u}_{2}{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\,\left({\rm{bracket}}!\right).\end{eqnarray} \tag{ 1.2 }$$

Using the inner product, the orthonormal property of the unit vectors can be expressed as

$$\begin{eqnarray}&&\langle {e}_{i}| {e}_{j}\rangle ={\delta }_{ij}\,{\rm{where}}\,{\delta }_{ij}=1\,{\rm{if}}\,i=j;{\rm{and}}\,0\,{\rm{if}}\,i\ne j\,\left(\mathrm{Kronecker}\mbox{'}\mathrm{s\; delta}\right).\end{eqnarray} \tag{ 1.3 }$$

Also, the definiton of the outer product of the two vectors is given by

$$\begin{eqnarray}&&| \mathit{\unicode[Book Antiqua]{x76}}\rangle \langle u| =\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right]\left[\begin{array}{cc}{u}_{1} & {u}_{2}\end{array}\right]=\left[\begin{array}{cc}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}{u}_{1} & {\mathit{\unicode[Book Antiqua]{x76}}}_{1}{u}_{2}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}{u}_{1} & {\mathit{\unicode[Book Antiqua]{x76}}}_{2}{u}_{2}\end{array}\right].\end{eqnarray} \tag{ 1.4 }$$

Matrices as operators on vectors

Because of the orthonormal property of the unit vectors, the vector components, $\mathit{\unicode[Book Antiqua]{x76}}$ ₁ and $\mathit{\unicode[Book Antiqua]{x76}}$ ₂, can be given by the inner products of the unit vector and the vector ∣$\mathit{\unicode[Book Antiqua]{x76}}$>:

$$\begin{eqnarray}&&\langle {e}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}\langle {e}_{1}| {e}_{1}\rangle +{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\langle {e}_{1}| {e}_{2}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1},\,{\rm{and}}\,\langle {e}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}\langle {e}_{2}| {e}_{1}\rangle +{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\langle {e}_{2}| {e}_{2}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{2}.\end{eqnarray} \tag{ 1.5 }$$

Therefore,

$$\begin{eqnarray}&&| \mathit{\unicode[Book Antiqua]{x76}}\rangle =\langle {e}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle | e{}_{1}\rangle +\langle {e}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle | {e}_{2}\rangle =| e{}_{1}\rangle \langle {e}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle +| {e}_{2}\rangle \langle {e}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle .\end{eqnarray} \tag{ 1.6 }$$

Here, the outer products of the unit vectors, $| e{}_{1}\rangle \langle {e}_{1}|$ and $| e{}_{2}\rangle \langle {e}_{2}|$, are called the projection operators:

$$\begin{eqnarray}{\hat{P}}_{1}=| e{}_{1}\rangle \langle {e}_{1}| =\left[\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right],{\rm{and}}\,{\hat{P}}_{2}=| e{}_{2}\rangle \langle {e}_{2}| =\left[\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right].\end{eqnarray} \tag{ 1.7 }$$

Notice ${\hat{P}}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}| {e}_{1}\rangle$, ${\hat{P}}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle ={\mathit{\unicode[Book Antiqua]{x76}}}_{2}| {e}_{2}\rangle$, ${\hat{P}}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle +{\hat{P}}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle =| \mathit{\unicode[Book Antiqua]{x76}}\rangle$, and thus

$$\begin{eqnarray}&&{\hat{P}}_{1}+{\hat{P}}_{2}=\hat{I}\end{eqnarray} \tag{ 1.8 }$$

where $\hat{I}$ is the 2×2 unit matrix.

Below is a summary of the addition and multiplication of matrices. Suppose

$$\begin{eqnarray*}A=\left[\begin{array}{cc}{a}_{11} & {a}_{12}\\ {a}_{21} & {a}_{22}\end{array}\right]\,\,{\rm{and}}\,B=\left[\begin{array}{cc}{b}_{11} & {b}_{12}\\ {b}_{21} & {b}_{22}\end{array}\right],\end{eqnarray*}$$

the basic matrix calculation rules are:

• (1)  
  $$\begin{eqnarray}\delta A=\left[\begin{array}{cc}\delta {a}_{11} & \delta {a}_{12}\\ \delta {a}_{21} & \delta {a}_{22}\end{array}\right]\,{\rm{where}}\,\delta \,{\rm{is}}\,{\rm{a}}\,{\rm{scalar}}\,{\rm{constant}},\end{eqnarray} \tag{ 1.9 }$$
• (2)  
  Addition:

  $$\begin{eqnarray}A\pm B=\left[\begin{array}{cc}{a}_{11}\pm {b}_{11} & {a}_{12}\pm {b}_{12}\\ {a}_{21}\pm {b}_{21} & {a}_{22}\pm {b}_{22}\end{array}\right],\end{eqnarray} \tag{ 1.10 }$$
• (3)  
  Multiplication:

$$\begin{eqnarray}AB=\left[\begin{array}{cc}{a}_{11}{b}_{11}+{a}_{12}{b}_{21} & {a}_{11}{b}_{12}+{a}_{12}{b}_{22}\\ {a}_{21}{b}_{11}+{a}_{22}{b}_{21} & {a}_{21}{b}_{12}+{a}_{22}{b}_{22}\end{array}\right].\end{eqnarray} \tag{ 1.11 }$$

Rotation and translation of a vector can be performed by applying corresponding matrices. For example, as shown in figure 1.1, rotating a vector ∣$\mathit{\unicode[Book Antiqua]{x76}}$> by angle θ to create a new vector, ∣$\mathit{\unicode[Book Antiqua]{x76}}$ '>, is expressed by ∣$\mathit{\unicode[Book Antiqua]{x76}}$ '>=R(θ)∣$\mathit{\unicode[Book Antiqua]{x76}}$>. The rotational matrix, R(θ), can be obtained as follows. Let ${\mathit{\unicode[Book Antiqua]{x76}}}_{1}=\mathit{\unicode[Book Antiqua]{x76}}\,\cos \,\alpha$ and ${\mathit{\unicode[Book Antiqua]{x76}}}_{2}=\mathit{\unicode[Book Antiqua]{x76}}\,\sin \,\alpha$. The components of the rotated vector are given by

$$\begin{eqnarray*}&&{\mathit{\unicode[Book Antiqua]{x76}}}_{1}^{^{\prime} }=\mathit{\unicode[Book Antiqua]{x76}}\,\cos (\alpha +\theta )=\mathit{\unicode[Book Antiqua]{x76}}\,\cos \alpha \,\cos \theta -\mathit{\unicode[Book Antiqua]{x76}}\,\sin \alpha \,\sin \theta ={\mathit{\unicode[Book Antiqua]{x76}}}_{1}\,\cos \theta -{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\,\sin \theta ,\end{eqnarray*}$$

and

$$\begin{eqnarray*}&&{\mathit{\unicode[Book Antiqua]{x76}}}_{2}^{^{\prime} }=\mathit{\unicode[Book Antiqua]{x76}}\,\sin (\alpha +\theta )=\mathit{\unicode[Book Antiqua]{x76}}\,\sin \alpha \,\cos \theta +\mathit{\unicode[Book Antiqua]{x76}}\,\cos \alpha \,\sin \theta ={\mathit{\unicode[Book Antiqua]{x76}}}_{2}\,\cos \theta +{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\,\sin \theta .\end{eqnarray*}$$

In the matrix representation (equation (1.11)), the above equations can be expressed as

$$\begin{eqnarray*}&&\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}^{\prime} \\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}^{\prime} \end{array}\right]=\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\,\cos \theta -{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\,\sin \theta \\ {\mathit{\unicode[Book Antiqua]{x76}}}_{1}\,\sin \theta +{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\,\sin \theta \end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\,\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right],\end{eqnarray*}$$

and the rotational matrix is, thus, given by

$$\begin{eqnarray}R(\theta )=\left[\begin{array}{cc}\cos \theta & -\,\sin \theta \\ \sin \theta & \cos \theta \end{array}\right].\end{eqnarray} \tag{ 1.12 }$$

Notice that rotation of the two-dimensional coordinate system (x, y) by angle θ to another coordinate system, (X, Y), is mathematically equivalent to the vector rotation by angle –θ. Thus,

$$\begin{eqnarray}&&\left[\begin{array}{c}X\\ Y\end{array}\right]=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\,\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right].\end{eqnarray} \tag{ 1.13 }$$

Consider two different two-dimensional vector spaces where each vector space has its own set of unit vectors or an orthonormal basis. Suppose a vector in each vector space is respectively given by

$$\begin{eqnarray*}&&| u\rangle =\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right]\,\,{\rm{and}}\,| \mathit{\unicode[Book Antiqua]{x76}}\rangle =\left[\begin{array}{c}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right].\end{eqnarray*}$$

A vector in the coupled vector space is expressed by the direct product of the two vectors:

$$\begin{eqnarray}&&| u\rangle \otimes | \mathit{\unicode[Book Antiqua]{x76}}\rangle =\left[\begin{array}{c}{u}_{1}| \mathit{\unicode[Book Antiqua]{x76}}\rangle \\ {u}_{2}| \mathit{\unicode[Book Antiqua]{x76}}\rangle \end{array}\right]=\left[\begin{array}{c}{u}_{1}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {u}_{2}{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\\ {u}_{2}{\mathit{\unicode[Book Antiqua]{x76}}}_{1}\\ {u}_{2}{\mathit{\unicode[Book Antiqua]{x76}}}_{2}\end{array}\right].\end{eqnarray} \tag{ 1.14 }$$

The coupled vector space becomes four-dimensional, and the coupled orthonormal basis is given by

$$\begin{eqnarray}&&| 0\rangle \otimes | 0\rangle =\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],| 0\rangle \otimes | 1\rangle =\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],| 1\rangle \otimes | 0\rangle =\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right],| 1\rangle \otimes | 1\rangle =\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right].\end{eqnarray} \tag{ 1.15 }$$

Notice that the direct product has the following distribution rule:

$$\begin{eqnarray}&&{\left(| {u}_{1}\rangle +| {u}_{2}\rangle \right)}_{A}\otimes {\left(| {\mathit{\unicode[Book Antiqua]{x76}}}_{1}\rangle +| {\mathit{\unicode[Book Antiqua]{x76}}}_{2}\rangle \right)}_{B}=| {u}_{1}{\rangle }_{A}| {\mathit{\unicode[Book Antiqua]{x76}}}_{1}{\rangle }_{B}+| {u}_{1}{\rangle }_{A}| {\mathit{\unicode[Book Antiqua]{x76}}}_{2}{\rangle }_{B}+| {u}_{2}{\rangle }_{A}| {\mathit{\unicode[Book Antiqua]{x76}}}_{1}{\rangle }_{B}+| {u}_{2}{\rangle }_{A}| {\mathit{\unicode[Book Antiqua]{x76}}}_{2}{\rangle }_{B},\end{eqnarray} \tag{ 1.16 }$$

where the subscripts, A and B, indicate two different vector spaces. Notice that we often write $| u\rangle \otimes | \mathit{\unicode[Book Antiqua]{x76}}\rangle =| u\rangle | \mathit{\unicode[Book Antiqua]{x76}}\rangle$ for short. It is important not to change the order of vector for the direct product.

We also define the direct product of two matrices, $A=\left[\begin{array}{cc}{a}_{11} & {a}_{12}\\ {a}_{21} & {a}_{22}\end{array}\right]$ and $B=\left[\begin{array}{cc}{b}_{11} & {b}_{12}\\ {b}_{21} & {b}_{22}\end{array}\right]$, as

$$\begin{eqnarray}A\otimes B=\left[\begin{array}{cc}{a}_{11}B & {a}_{12}B\\ {a}_{21}B & {a}_{22}B\end{array}\right]=\left[\begin{array}{cccc}{a}_{11}{b}_{11} & {a}_{11}{b}_{12} & {a}_{12}{b}_{11} & {a}_{12}{b}_{12}\\ {a}_{11}{b}_{21} & {a}_{11}{b}_{22} & {a}_{12}{b}_{21} & {a}_{12}{b}_{22}\\ {a}_{21}{b}_{11} & {a}_{21}{b}_{12} & {a}_{22}{b}_{11} & {a}_{22}{b}_{12}\\ {a}_{21}{b}_{21} & {a}_{21}{b}_{22} & {a}_{22}{b}_{21} & {a}_{22}{b}_{22}\end{array}\right].\end{eqnarray} \tag{ 1.17 }$$

We will use the direct products when we discuss universal operators in chapter 2.

Here is a summary of properties that quantum systems must satisfy [1]. Whenever we create a quantum algorithm, it must satisfy these conditions.

• (1)  
  The quantum states of a system can be described by a single-valued continuous complex-valued function, $| \psi \rangle$, and the physics observables, A, including energy, position, momentum, angular momentum, spins, and particle creation/annihilation, can be expressed as a mathematical operator, $\hat{A}$. Measurement values, {ε_k ; k=1, 2, ...}, of the observable A satisfy the equation, $\hat{A}| {\psi }_{k}\rangle ={\varepsilon }_{k}| {\psi }_{k}\rangle$, where ε_k is the eigen value, and $| \psi {}_{k}\rangle$ is eigen function of the equation. When the observable is Hamiltonian (energy), the equation is called the Schrödinger equation, $\hat{H}| {\psi }_{k}\rangle ={\varepsilon }_{k}| {\psi }_{k}\rangle$.
• (2)  
  A single measurement of the physical observable A yields one of the possible eigen values {ε_k }. The equation itself does not determine the eigen functions and the eigen values without a boundary condition.
• (3)  
  A quantum state can be normalized, and its operators are linear:

  • (i)  
    $\langle \psi | \psi \rangle =1$ where $\langle \psi | ={\left(| \psi \rangle \right)}^{\dagger }$ is the conjugate transpose of ∣ψ>, and
  • (ii)  
    $\hat{A}\left(a| {\psi }_{1}\rangle +b| {\psi }_{2}\rangle \right)=a\hat{A}| {\psi }_{1}\rangle +b\hat{A}| {\psi }_{2}\rangle$ where the coefficients a and b are constants of complex values.
• (4)  
  Because the physical observables are real values, the eigen values must be real numbers. Therefore, the operator $\hat{A}$ must be Hermitian, i.e., ${\hat{A}}^{\dagger }=\hat{A}$.Proof: Suppose $\hat{A}| {\psi }_{k}\rangle ={\varepsilon }_{k}| {\psi }_{k}\rangle$, and the eigen value is real, i.e., ${\varepsilon }_{k}={\varepsilon }_{k}^{* }$. $\langle {\psi }_{k}| \hat{A}| {\psi }_{k}\rangle =\langle {\psi }_{k}| {\varepsilon }_{k}| {\psi }_{k}\rangle ={\varepsilon }_{k}\langle {\psi }_{k}| {\psi }_{k}\rangle ={\varepsilon }_{k},{\rm{a}}{\rm{n}}{\rm{d}}$ ${\left(\langle {\psi }_{k}| \hat{A}| {\psi }_{k}\rangle \right)}^{\dagger }=\langle {\psi }_{k}| {\hat{A}}^{\dagger }| {\psi }_{k}\rangle ={\varepsilon }_{k}^{* }$.Thus, if the eigenvalue is a real number, ${\hat{A}}^{\dagger }=\hat{A}$.
• (5)  
  The eigen functions are orthogonal, i.e., $\langle {\psi }_{i}| {\psi }_{j}\rangle =0$ if $i\ne j$.Proof. Suppose $\hat{A}| {\psi }_{i}\rangle ={\varepsilon }_{i}| {\psi }_{i}\rangle$ and $\hat{A}| {\psi }_{j}\rangle ={\varepsilon }_{j}| {\psi }_{j}\rangle$ where $\hat{A}$ is Hermitian.Notice ${\left(\hat{A}| {\psi }_{i}\rangle \right)}^{\dagger }=\langle {\hat{A}}^{\dagger }{\psi }_{i}| =\langle \hat{A}{\psi }_{i}|$ because $\hat{A}$ is Hermitian. Also notice that because the eigen values are real, ${\left(\hat{A}| {\psi }_{i}\rangle \right)}^{\dagger }={\left({\varepsilon }_{i}| {\psi }_{i}\rangle \right)}^{\dagger }={\varepsilon }_{i}\langle {\psi }_{i}|$. Therefore, $\langle {\hat{A}}^{\dagger }{\psi }_{i}| {\psi }_{j}\rangle =\langle {\psi }_{i}| \hat{A}{\psi }_{j}\rangle ={\varepsilon }_{j}\langle {\psi }_{i}| {\psi }_{j}\rangle ,{\rm{a}}{\rm{n}}{\rm{d}}$ $\langle {\hat{A}}^{\dagger }{\psi }_{i}| {\psi }_{j}\rangle -\langle {\psi }_{i}| \hat{A}{\psi }_{j}\rangle =\left({\varepsilon }_{i}-{\varepsilon }_{j}\right)\langle {\psi }_{i}| {\psi }_{j}\rangle =0$. Thus, $\langle {\psi }_{i}| {\psi }_{j}\rangle =0$ unless ${\varepsilon }_{i}={\varepsilon }_{j}$.
• (6)  
  The complete set of eigen functions {$| \psi {}_{k}\rangle$, k=1, 2, 3, ....} forms an orthonormal basis. In other words, any quantum state of the given system can be expressed by the superposition of eigen functions: $| \psi \rangle =\displaystyle \displaystyle \sum _{k}{c}_{k}| {\psi }_{k}\rangle$ where $\langle {\psi }_{j}| {\psi }_{k}\rangle ={\delta }_{jk}$ and ${c}_{k}=\langle {\psi }_{k}| \psi \rangle$.
• (7)  
  Suppose a coordinate system for observation is changed to another coordinated system, a complete set of the eigen functions {$| \psi {}_{k}\rangle$, k=1, 2, 3, ....} is transformed to another set of eigen functions {$| \varphi {}_{k}\rangle$, k=1, 2, 3, ....}, i.e., $| \varphi \rangle =\hat{U}| \psi \rangle$ where $\hat{U}$ is the operator for this transformation. Then, $| \varphi {\rangle }_{j}=\displaystyle \displaystyle \sum _{k=1}{u}_{jk}| \psi {\rangle }_{k}$, and $\langle {\varphi }_{i}| {\varphi }_{j}\rangle =\displaystyle \displaystyle \sum _{k,m}{({u}_{ik})}^{\dagger }{u}_{mj}\langle {\psi }_{k}| {\psi }_{m}\rangle$. Since both sets of eigen functions are orthonormal, $\langle {\varphi }_{i}| {\varphi }_{j}\rangle ={\delta }_{i,j}=\displaystyle \displaystyle \sum _{k,m}{({u}_{ik})}^{\dagger }{u}_{mj}{\delta }_{k,m}$.Note: An operator, which satisfies $\displaystyle \displaystyle \sum _{k}{({u}_{ik})}^{\dagger }{u}_{kj}={\delta }_{i,j}$, is called the unitary operator, and ${\hat{U}}^{\dagger }\hat{U}=\hat{I}$, i.e., ${\hat{U}}^{\dagger }={\hat{U}}^{-1}$. Operators that change quantum states must be unitary.
• (8)  
  The expectation value of the observable A is given by $\langle A\rangle =\langle \psi | \hat{A}| \psi \rangle$.In particular, $\langle A\rangle =\langle {\psi }_{k}| \hat{A}| {\psi }_{k}\rangle ={\varepsilon }_{k}$ if $| \psi \rangle =| {\psi }_{k}\rangle$.
• (9)  
  A projection operator is given by ${\hat{P}}_{k}=| {\psi }_{k}\rangle \langle {\psi }_{k}|$.If $| \psi \rangle =\displaystyle \sum _{k}{c}_{k}| {\psi }_{k}\rangle$, then ${\hat{P}}_{k}| \psi \rangle =\displaystyle \sum _{j}{c}_{j}| {\psi }_{k}\rangle \langle {\psi }_{k}| {\psi }_{j}\rangle =\displaystyle \sum _{j}{c}_{j}| {\psi }_{k}\rangle {\delta }_{kj}={c}_{k}| {\psi }_{k}\rangle$.The probability of observing the eigen state $| \psi {}_{k}\rangle$ is

  $$\begin{eqnarray*}&&| \langle \psi | {\hat{P}}_{k}| \psi \rangle {| }^{2}=| \langle \psi | \displaystyle \sum _{j}{c}_{j}| {\psi }_{k}\rangle \langle {\psi }_{k}| {\psi }_{j}\rangle {| }^{2}=| \langle \psi | \displaystyle \sum _{j}{c}_{j}| {\psi }_{k}\rangle {\delta }_{kj}{| }^{2}=| {c}_{k}\langle \psi | {\psi }_{k}\rangle {| }^{2}=| {c}_{k}{| }^{2}.\end{eqnarray*}$$
• (10) Matrix representation of operator $\hat{U}$.

Let $| \psi \rangle =\displaystyle \displaystyle \sum _{j}{c}_{j}| {\psi }_{k}\rangle$ where ${c}_{j}=\langle {\psi }_{j}| \psi \rangle$, and $| \psi ^{\prime} \rangle =\hat{U}| \psi \rangle =\displaystyle \sum _{j}{c}_{j}\hat{U}| {\psi }_{j}\rangle$.

If $| \psi ^{\prime} \rangle =\displaystyle \displaystyle \sum _{j}c{^{\prime} }_{j}| {\psi }_{k}\rangle$ where $c{^{\prime} }_{j}=\langle {\psi }_{j}| \psi ^{\prime} \rangle$, then $| \psi ^{\prime} \rangle$ becomes

$| \psi ^{\prime} \rangle =\displaystyle \sum _{i}c^{\prime} i| {\psi }_{i}\rangle =\displaystyle \sum _{i}{c}_{i}\hat{U}| {\psi }_{i}\rangle$, and thus,

$$\begin{eqnarray*}&&c{^{\prime} }_{i}=\langle {\psi }_{i}| \psi ^{\prime} \rangle =\langle {\psi }_{i}| \hat{U}\psi \rangle =\langle {\psi }_{i}| \displaystyle \sum _{j}{c}_{j}\hat{U}{\psi }_{j}\rangle =\displaystyle \sum _{j}\langle {\psi }_{i}| \hat{U}| {\psi }_{j}\rangle {c}_{j}.\end{eqnarray*}$$

Define $\langle {\psi }_{i}| \hat{U}| {\psi }_{j}\rangle ={U}_{ij}$, and we obtain $c^{\prime} {}_{i}=\displaystyle \sum _{j}{U}_{ij}{c}_{j}$. Therefore, the transform

$| \psi \rangle \to | \psi ^{\prime} \rangle =\hat{U}| \psi \rangle$ can be represented by a linear transform of {c_i}: $c{^{\prime} }_{i}=\displaystyle \sum _{j}{U}_{ij}{c}_{j}$.

An electron is known to have two possible spin states. It can be described on a two-dimensional complex vector space. When we measure the spin state along the z-axis or the 'vertical' axis, which is the default direction of external magnetic field, we observe the spin up (↑) or down (↓) state. We assign the following vectors ∣0> and ∣1> to the spin states:

$$\begin{eqnarray}&&\begin{array}{c}{\rm{S}}{\rm{p}}{\rm{i}}{\rm{n}}\,{\rm{u}}{\rm{p}}:| \uparrow \rangle =| 0\rangle =\left[\begin{array}{c}1\\ 0\end{array}\right],\,{\rm{a}}{\rm{n}}{\rm{d}}\,{\rm{s}}{\rm{p}}{\rm{i}}{\rm{n}}\,{\rm{d}}{\rm{o}}{\rm{w}}{\rm{n}}:| \downarrow \rangle =| 1\rangle =\left[\begin{array}{c}0\\ 1\end{array}\right].\end{array}\end{eqnarray} \tag{ 1.18 }$$

The spin state vectors form an orthonormal basis of the vector space, {∣0>, ∣1>} where $\langle 0| 0\rangle =\langle 1| 1\rangle =1$, and $\langle 0| 1\rangle =\langle 1| 0\rangle =0$. An arbitrary spin state can be given by $| \psi \rangle =a| 0\rangle +b| 1\rangle$ where the coefficients a and b are complex value constants, and the spin state is normalized, $\left|\langle \psi | \psi \rangle \right|=| a{| }^{2}+| b{| }^{2}=1$ because either spin state will be observed by a measurement. The spin state $| \psi \rangle$ is called a qbit in the quantum computation/information. The complex coefficients, a and b, can be expressed as $a=\,\cos \,\left(\displaystyle \frac{\theta }{2}\right)$ and $b={e}^{i\varphi }\,\sin \,\left(\displaystyle \frac{\theta }{2}\right)$, using phase angles, φ and θ, which satisfy the normalization condition, ∣a∣² + ∣b∣²=1. A Bloch sphere, shown in figure 1.2, is often used to visualize the qbit dynamics although we will not use it in this book.

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If one observes the spin state, $| \psi \rangle =a| 0\rangle +b| 1\rangle$, along the z-axis, the measurement yields either spin up or down. Observation of the spin state can be interpreted as the application of the projection operators. From equation (1.7), the projection operators for the spin state described by the orthonormal basis, {∣0>, ∣1>}, are given

$$\begin{eqnarray}&&{\hat{P}}_{0}=| 0\rangle \langle 0| =\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{cc}1 & 0\end{array}\right]=\left[\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right]\,\,{\rm{and}}\,{\hat{P}}_{1}=| 1\rangle \langle 1| =\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}0 & 1\end{array}\right]=\left[\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right].\end{eqnarray} \tag{ 1.19 }$$

The probability of observing the spin up state along the z-axis, ∣a∣², can be calculated from a measurement

$$\begin{eqnarray}{\hat{P}}_{0}| \psi \rangle =| 0\rangle \langle 0| \psi \rangle =| 0\rangle \langle 0| \left(a| 0\rangle +b| 1\rangle \right)=a\left[\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]+b\left[\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=a\left[\begin{array}{c}1\\ 0\end{array}\right]=a| 0\rangle ,\end{eqnarray} \tag{ 1.20 }$$

and the probability of observing the spin down state along the z-axis, ∣b∣², can be calculated from a measurement

$$\begin{eqnarray}{\hat{P}}_{1}| \psi \rangle =| 1\rangle \langle 1| \psi \rangle =| 1\rangle \langle 1| \left(a| 0\rangle +b| 1\rangle \right)=a\left[\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]+b\left[\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=b\left[\begin{array}{c}0\\ 1\end{array}\right]=b| 1\rangle .\end{eqnarray} \tag{ 1.21 }$$

Once a measurement is conducted on a quantum system, the quantum state of the system is collapsed, and a successive measurement yields the same state as the previous measurement. This means

$$\begin{eqnarray*}&&{\hat{P}}_{0}{\hat{P}}_{0}={\hat{P}}_{0},\,{\hat{P}}_{1}{\hat{P}}_{1}={\hat{P}}_{1},\,{\rm{and}}\,{\hat{P}}_{0}{\hat{P}}_{1}={\hat{P}}_{1}{\hat{P}}_{0}=0.\end{eqnarray*}$$

To obtain the time evolution and rotation of spins, we use exponential operators. We define the form of exponential operators as

$$\begin{eqnarray}&&{e}^{u\hat{A}}=\exp (u\hat{A})\equiv \displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{1}{n!}{\left(u\hat{A}\right)}^{n}\,=\,\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{u}^{n}}{n!}{\hat{A}}^{n}\end{eqnarray} \tag{ 1.22 }$$

where u is a complex number. Suppose the wave function is given by $| \psi (0)\rangle$ at t = 0, the wave function satisfies the time dependent Schrödinger equation:

$$\begin{eqnarray*}&&i\hslash \displaystyle \frac{\partial | \psi (t)\rangle }{\partial t}=\hat{H}| \psi (t)\rangle \end{eqnarray*}$$

where $\hat{H}$ is the Hamiltonian of the system. From the Schrödinger equation, the time evolution of the wave function is given by

$$\begin{eqnarray}&&| \psi (t)\rangle =\exp \left(-\displaystyle \frac{i}{\hslash }\hat{H}t\right)| \psi (0)\rangle .\end{eqnarray} \tag{ 1.23 }$$

Proof.

$$\begin{eqnarray*}&&\begin{array}{l}i\hslash \displaystyle \frac{\partial | \psi (t)\rangle }{\partial t}=\displaystyle \frac{\partial }{\partial t}\left({e}^{-\displaystyle \frac{i}{\hslash }\hat{H}t}| \psi (0)\rangle \right)=i\hslash \displaystyle \frac{\partial }{\partial t}\left[\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{1}{n!}{\left(-\displaystyle \frac{i\hat{H}t}{\hslash }\right)}^{n}\right]| \psi (0)\rangle \\ =\,\hat{H}\displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{1}{(n-1)!}{\left(-\displaystyle \frac{i\hat{H}t}{\hslash }\right)}^{n-1}(n-1){t}^{n-1}| \psi (0)\rangle \\ =\,i\hslash \displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{1}{n!}{\left(-\displaystyle \frac{i\hat{H}t}{\hslash }\right)}^{n}(n{t}^{n-1})| \psi (0)\rangle \\ =\,-i\hslash \left(\displaystyle \frac{i\hat{H}}{\hslash }\right)\displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{1}{(n-1)!}{\left(-\displaystyle \frac{i\hat{H}t}{\hslash }\right)}^{n-1}(n-1){t}^{n-1}| \psi (0)\rangle \\ =\,\hat{H}{e}^{-\displaystyle \frac{i}{\hslash }\hat{H}t}| \psi (0)\rangle =\hat{H}| \psi (t)\rangle .\end{array}\end{eqnarray*}$$

Define the rotational operator, ${\hat{R}}_{j}(\delta \theta )$ of an infinitesimal angle $\delta \vec{\theta }$ about the rotational axis j, such that ${\hat{R}}_{j}(\delta \vec{\theta })| \psi (\vec{r})\rangle =| \psi ^{\prime} (\vec{r})\rangle$ on the wave function $| \psi (\vec{r})\rangle$. Because the position of the vector after it is rotated forward by an infinitesimal angle $\delta \vec{\theta }$ is given by $\vec{r}^{\prime} =\vec{r}+\delta \vec{\theta }\times \vec{r}$ using the vector product (figure 1.3), we obtain

$$\begin{eqnarray*}&&| \psi ^{\prime} (\vec{r})\rangle =| \psi (\vec{r}-\delta \vec{\theta }\times \vec{r})\rangle ,\end{eqnarray*}$$

and thus,

$$\begin{eqnarray}&&\begin{array}{l}{\hat{R}}_{j}(\delta \vec{\theta })| \psi (\vec{r})\rangle =| \psi (\vec{r}-\delta \vec{\theta }\times \vec{r})\rangle \approx | \psi (\vec{r})\rangle -\left(\delta \vec{\theta }\times \vec{r}\right)\cdot \nabla | \psi (\vec{r})\rangle \\ =\,| \psi (\vec{r})\rangle -\displaystyle \frac{i}{\hslash }\left(\delta \vec{\theta }\times \vec{r}\right)\cdot \vec{p}| \psi (\vec{r})\rangle =\left(1-\displaystyle \frac{i}{\hslash }\delta \vec{\theta }\cdot \vec{L}\right)| \psi (\vec{r})\rangle \end{array}\end{eqnarray} \tag{ 1.24 }$$

where $\vec{p}=-i\hslash \nabla$ is the linear momentum, and $\vec{L}=\vec{r}\times \vec{p}$ is the angular momentum.

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The infinitesimal angle is defined as $\delta \vec{\theta }=\mathop{\mathrm{lim}}\limits_{N\to \infty }\,\vec{\theta }/N$, and we obtain

$$\begin{eqnarray}&&{\hat{R}}_{j}(\vec{\theta })=\mathop{\mathrm{lim}}\limits_{N\to \infty }\,{\left(1-\displaystyle \frac{i}{\hslash }\displaystyle \frac{\vec{\theta }\cdot \vec{L}}{N}\right)}^{N}=\exp (-i\vec{\theta }\cdot \vec{L}/\hslash ).\end{eqnarray} \tag{ 1.25 }$$

The spin rotational operators about the x, the y, and the z axes are, respectively, given by replacing the angular momentum of each axis with Pauli's spin matrix of the corresponding axis:

$$\begin{eqnarray}&&{\hat{R}}_{j}(\vec{\theta })=\exp (-i\displaystyle \frac{{\theta }_{j}{\hat{L}}_{j}}{\hslash })=\exp (-i\displaystyle \frac{{\theta }_{j}{\hat{\sigma }}_{j}}{2}),\end{eqnarray} \tag{ 1.26 }$$

where j = x, y, or z, and

${\hat{\sigma }}_{x}=\left[\begin{array}{cc}0 & 1\\ 1 & 0\end{array}\right]$, ${\hat{\sigma }}_{y}=\left[\begin{array}{cc}0 & -\,i\\ i & 0\end{array}\right]$, ${\hat{\sigma }}_{z}=\left[\begin{array}{cc}1 & 0\\ 0 & -\,1\end{array}\right]$, and $\hat{I}=\left[\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right]$.

Thus,

$$\begin{eqnarray}{\hat{R}}_{x}(\theta )=\exp (-i\displaystyle \frac{\theta {\hat{\sigma }}_{x}}{2})=\left(\cos \,\displaystyle \frac{\theta }{2}\right)\hat{I}+i\left(\sin \,\displaystyle \frac{\theta }{2}\right){\hat{\sigma }}_{x}=\left[\begin{array}{cc}\cos \left(\theta /2\right) & -\,i\,\sin \left(\theta /2\right)\\ -\,i\,\sin \left(\theta /2\right) & \cos \left(\theta /2\right)\end{array}\right],\end{eqnarray} \tag{ 1.27 }$$

$$\begin{eqnarray}{\hat{R}}_{y}(\theta )=\exp (-i\displaystyle \frac{\theta {\hat{\sigma }}_{y}}{2})=\left(\cos \,\displaystyle \frac{\theta }{2}\right)\hat{I}+i\left(\sin \,\displaystyle \frac{\theta }{2}\right){\hat{\sigma }}_{y}=\left[\begin{array}{cc}\cos \,\left(\theta /2\right) & -\,\sin \,\left(\theta /2\right)\\ \sin \,\left(\theta /2\right) & \cos \,\left(\theta /2\right)\end{array}\right],\end{eqnarray} \tag{ 1.28 }$$

$$\begin{eqnarray}{\hat{R}}_{z}(\theta )=\exp (-i\displaystyle \frac{\theta {\hat{\sigma }}_{z}}{2})=\left(\cos \,\displaystyle \frac{\theta }{2}\right)\hat{I}+i\left(\sin \,\displaystyle \frac{\theta }{2}\right){\hat{\sigma }}_{z}=\left[\begin{array}{cc}{e}^{-i\theta /2} & 0\\ 0 & {e}^{i\theta /2}\end{array}\right].\end{eqnarray} \tag{ 1.29 }$$

where we have omitted the suffix of the angles. For basic concepts of spin states, refer to a textbook on quantum mechanics [1].

In order to observe a spin state, an external magnetic field is applied to the spin. If the external magnet is along the z-axis, we observe either ∣0> or ∣1> as we discussed above (equations (1.19) and (1.20)). Now, if we rotate the external magnetic field by angle θ about one of the coordinate axes, we rotate the observation coordinate frame. Thus, if we rotate the z-axis by angle θ on the zx-plane, i.e., if we rotate the 'vertical' spin-measurement coordinate frame about the y-axis by angle θ, we obtain a new 'tilted' measurement coordinate frame where the rotated orthonormal basis is given by:

$$\begin{eqnarray}| 0^{\prime} \rangle ={\hat{R}}_{y}(-\theta )| 0\rangle =\left[\begin{array}{cc}\cos (\theta /2) & \sin (\theta /2)\\ -\,\sin (\theta /2) & \cos (\theta /2)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}\cos (\theta /2)\\ -\,\sin (\theta /2)\end{array}\right],\end{eqnarray} \tag{ 1.30 }$$

and

$$\begin{eqnarray*}| 1^{\prime} \rangle ={\hat{R}}_{y}(-\theta )| 1\rangle =\left[\begin{array}{cc}\cos (\theta /2) & \sin (\theta /2)\\ -\,\sin (\theta /2) & \cos (\theta /2)\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}\sin (\theta /2)\\ \cos (\theta /2)\end{array}\right].\end{eqnarray*}$$

In particular, if θ=π/2, the direction of the external magnetic field becomes 'horizontal,' or

$$\begin{eqnarray}&&| 0^{\prime} \rangle =| \to \rangle =\left[\begin{array}{c}\cos (\pi /4)\\ -\,\sin (\pi /4)\end{array}\right]=\left[\begin{array}{c}1/\sqrt{2}\\ -\,1/\sqrt{2}\end{array}\right],\,{\rm{and}}\,| 1^{\prime} \rangle =| \leftarrow \rangle =\left[\begin{array}{c}\sin (\pi /4)\\ \cos (\pi /4)\end{array}\right]=\left[\begin{array}{c}1/\sqrt{2}\\ 1/\sqrt{2}\end{array}\right].\end{eqnarray} \tag{ 1.31 }$$

Recall that if we observe the spin state, $| \psi \rangle =a| 0\rangle +b| 1\rangle$, along the 'vertical' coordinates frame where the external magnetic field is along the z-axis, an observed state is either ∣0> or ∣1> with the corresponding probability ∣a∣² or ∣b∣². If we observe the spin state using the 'horizontal' measurement frame where the external magnetic field is along the x-axis, the spin state to be observed can be determined by applying the projection operators, ${\hat{P}}_{| 0^{\prime} \rangle }=| 0^{\prime} \rangle \langle 0^{\prime} |$ and ${\hat{P}}_{| 1^{\prime} \rangle }=| 1^{\prime} \rangle \langle 1^{\prime} |$ to the basis ∣0> and ∣1>:

$$\begin{eqnarray}{\hat{P}}_{| 0^{\prime} \rangle }| 0\rangle =a(| 0^{\prime} \rangle \langle 0^{\prime} | )| 0\rangle =a\left[\begin{array}{cc}1/\sqrt{2} & 1/\sqrt{2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]| 0^{\prime} \rangle =\displaystyle \frac{a}{\sqrt{2}}| 0^{\prime} \rangle ,\end{eqnarray} \tag{ 1.32 }$$

and

$$\begin{eqnarray*}{\hat{P}}_{| 1^{\prime} \rangle }| 1\rangle =b(| 1^{\prime} \rangle \langle 1^{\prime} | )| 1\rangle =b\left[\begin{array}{cc}1/\sqrt{2} & -\,1/\sqrt{2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]| 1^{\prime} \rangle =\displaystyle \frac{b}{\sqrt{2}}| 1^{\prime} \rangle .\end{eqnarray*}$$

Thus, the spin state becomes $| \psi \rangle =\displaystyle \frac{1}{\sqrt{2}}(a| 0^{\prime} \rangle +b| 1^{\prime} \rangle )$ in the 'horizontal' measurement frame, and we observe ∣0'> or ∣1'>, and the probability of observing ∣0'> is ∣a∣²/2 and the probability of observing ∣1'> spin is ∣b∣²/2 along the horizontal magnetic field. This is a useful description when we discuss Bell's inequality where three different observation frames are involved (section 5.1).

Suppose a set of two electron spins, spin A and spin B, are coupled, the net spin S is 1 or 0. Such an example is two electrons in a hydrogen molecule. As shown in table 1.1, if S = 1, then S_z = +1, 0, or −1 whereas, if S=0, then S_z = 0 only.

Table 1.1.  Two coupled spins.

Net spin S	Sz	Quantum state
S = 1	+1	$| {0}_{A}{0}_{B}\rangle$
	0	$\displaystyle \frac{1}{\sqrt{2}}\left(| {0}_{A}{1}_{B}\rangle +| {0}_{A}{1}_{B}\rangle \right)$
	−1	$| {1}_{A}{1}_{B}\rangle$
S = 0	0	$\displaystyle \frac{1}{\sqrt{2}}\left(| {0}_{A}{1}_{B}\rangle -| {1}_{A}{0}_{B}\rangle \right)$

The quantum states of S=1 and S_z = 0 or S= 0 and S_z = 0 are called the entangled states. Suppose we conduct an observation of spin A when S = 0. Using the projection operator, the spin state becomes

$$\begin{eqnarray}&&{\hat{P}}_{A}| \psi \rangle =| {0}_{A}\rangle \langle {0}_{A}| \left(\displaystyle \frac{1}{\sqrt{2}}\left(| {0}_{A}{1}_{B}\rangle -| {1}_{A}{0}_{B}\rangle \right)\right)=\displaystyle \frac{1}{\sqrt{2}}| {0}_{A}{1}_{B}\rangle .\end{eqnarray} \tag{ 1.34 }$$

That is, if spin A is up (↑), then spin B must be down (↓). Observing the spin A will automatically determine the spin state B. It seems to be trivial, and this is also true for the classical physics. However, recall that unless we observe one of them, both spins A and B have equal probabilities of spin up and down!

What is important to perceive is that, in classical physics, all spin states are pre-determined regardless of whether we observe them or not. Therefore, the spin states A and B are already determined with the condition of the total spin value. On the other hand, in quantum physics, unless we make observation of spin A, spin A can be both up and down with equal probabilities, and spin B can also be both up and down with equal probabilities. This argument is very much like the quantum logic of Schrödinger's cat.

According to quantum mechanics, an act of observation of the spin A state instantly determines the spin B state, depending on the outcome of the spin state A. Remember that the spin A state is undermined before observation. This is why we call it the 'spooky' behavior of quantum entanglement. Furthermore, the entanglement is non-localizing, i.e., even if the spins are physically separated (without observation) by an astronomical distance after making the coupled state, they are still entangled!

The quantum entanglement is such a strange and unique quantum behavior, and it is one of the key properties that make quantum computation and information theory distinct from the classical theory. One of its astonishing examples is quantum teleportation, which will be discussed in chapter 5.

Imagine that two classical sinusoidal waves of frequencies, sin(ω₁ t) and sin(ω₂ t) are superposed, the resultant wave is given by

$$\begin{eqnarray*}&&\sin ({\omega }_{1}t)+\,\sin ({\omega }_{2}t)=2\,\cos \,\left(\displaystyle \frac{{\omega }_{1}-{\omega }_{2}}{2}t\right)\,\sin \,\left(\displaystyle \frac{{\omega }_{1}+{\omega }_{2}}{2}t\right).\end{eqnarray*}$$

There are two possible frequencies after superposition of two classical waves. The quantum superposition is fundamentally different from the superposition of n-classical waves which linearly gives only n different states. The number of states exponentially increases in the quantum superposition.

A qbit $| \psi \rangle =a| 0\rangle +b| 1\rangle$ is a superposition of ∣0> and ∣1> states. The possible states from superposition of two qbits are given by the direct product defined by equation (1.14)

$$\begin{eqnarray}&&| \psi \rangle =\left(a| 0\rangle +b| 1\rangle \right)\otimes \left(c| 0\rangle +d| 1\rangle \right)=ac| 00\rangle +ad| 01\rangle +bc| 10\rangle +bd| 11\rangle =\left[\begin{array}{c}ac\\ ad\\ bc\\ bd\end{array}\right].\end{eqnarray} \tag{ 1.35 }$$

Now, the number of possible states become four. If n-qbits are superposed,

$$\begin{eqnarray}&&\begin{array}{l}| \psi \rangle \otimes | \psi \rangle \otimes \cdots \otimes | \psi \rangle \\ =\,\left(\displaystyle \frac{1}{\sqrt{2}}| 0\rangle +\displaystyle \frac{1}{\sqrt{2}}| 1\rangle \right)\otimes \left(\displaystyle \frac{1}{\sqrt{2}}| 0\rangle +\displaystyle \frac{1}{\sqrt{2}}| 1\rangle \right)\otimes \cdots \,\otimes \,\left(\displaystyle \frac{1}{\sqrt{2}}| 0\rangle +\displaystyle \frac{1}{\sqrt{2}}| 1\rangle \right)\\ =\,\displaystyle \frac{1}{\sqrt{{2}^{n}}}\left(| 0\rangle \otimes | 0\rangle \otimes \cdots \,\otimes | 0\rangle +\cdots \cdots +| 1\rangle \otimes | 1\rangle \otimes \cdots \,\otimes \,| 1\rangle \right)\\ =\,\displaystyle \frac{1}{\sqrt{{2}^{n}}}\left(| 0\rangle | 0\rangle \cdots | 0\rangle +\cdots \cdots +| 1\rangle | 1\rangle \cdots | 1\rangle \right)=\displaystyle \frac{1}{\sqrt{{2}^{n}}}\left(| 00\rangle \cdots 0\rangle +\cdots \cdots +| \mathrm{11\cdots 1}\rangle \right).\end{array}\end{eqnarray} \tag{ 1.36 }$$

This represents a superposed 2ⁿ -state from $| \mathrm{00\cdots 0}\rangle$ to $| \mathrm{11\cdots 1}\rangle$ with equal weight. Computation using superposed qbits is superior to the classical binary bit calculation (quantum supremacy) because we can perform computation on all the superposed states at once.

It must be noted that the quantum entanglement is not superposition of quantum states. If two qbits are entangled, they cannot be expressed as the direct product of two qbits. In other words, they cannot be separated as two independent quantum events. In chapter 3, we show quantum gates that superpose, entangle, and detangle qbits using conditional gates.