Review of Linear Algebra in Quantum Mechanics (V)

More about Orthonormal Basis

First of all, consider any vector $|\psi\rangle$ expanded within an orthonormal basis $|\psi\rangle=\sum_{i=1}^n\psi_i|\epsilon_i\rangle$. Because of the orthonormal property \begin{equation}\langle \epsilon_i |\epsilon_j\rangle=\delta_{ij}\,,\tag{1}\end{equation}
by applying $\langle\epsilon_i|$ on both sides of the expansion , we obtain $\psi_i = \langle \epsilon_i |\psi\rangle$. Since the relation \begin{equation}|\psi\rangle =\sum_{i=1}^n\psi_i|\epsilon_i\rangle = \sum_{i=1}^n \langle \epsilon_i |\psi\rangle\,|\epsilon_i\rangle = \sum_{i=1}^n |\epsilon_i\rangle\langle\epsilon_i |\psi\rangle\end{equation} holds for any $|\psi\rangle$, we obtain the completeness property \begin{equation} \sum_{i=1}^n |\epsilon_i\rangle\langle\epsilon_i |= \cal{I}\,,\tag{2}\end{equation} where $\cal{I}$ is the identity transformation that leaves any vector unchanged. 
 
Secondly, let's write an operator within the orthonormal basis \begin{equation}  \cal{O}\, \Big[|\epsilon_1\rangle\,\,|\epsilon_2\rangle\,\,\cdots\,\,| \epsilon_n\rangle\Big]=\Big[|\epsilon_1\rangle\,\,|\epsilon_2\rangle\,\,\cdots\,\,|\epsilon_n\rangle\Big]\begin{bmatrix} o_{11} & o_{12} & \cdots & o_{1n} \\ o_{21} & o_{22} & \cdots & o_{2n} \\ \vdots  & \vdots  & \ddots & \vdots  \\ o_{n1} & o_{n2} & \cdots & o_{nn} \end{bmatrix}\,,\tag{3}\end{equation} 
  1. if we multiply $\left[\begin{array}{c}\langle \epsilon_1| \\ \vdots \\ \langle \epsilon_n| \end{array}\right]$ from the left on both side of Eq. (3) and use the orthonormal property (1), we obtain \begin{equation}\left[\begin{array}{c}\langle \epsilon_1| \\ \langle \epsilon_2 \\ \vdots \\ \langle \epsilon_n| \end{array}\right]\cal{O}\, \Big[|\epsilon_1\rangle\,\,|\epsilon_2\rangle\,\,\cdots\,\,| \epsilon_n\rangle\Big]=\begin{bmatrix} o_{11} & o_{12} & \cdots & o_{1n} \\ o_{21} & o_{22} & \cdots & o_{2n} \\ \vdots  & \vdots  & \ddots & \vdots  \\ o_{n1} & o_{n2} & \cdots & o_{nn} \end{bmatrix}\,,\end{equation} that is \begin{equation} o_{ij}=\langle \epsilon_i | \cal{O}|\epsilon_j\rangle\,.\tag{4}\end{equation} The matrix element of an operator within the orthonormal basis can be obtained simply by multiplying the corresponding bra and ket from the two sides of the operator.
  2. if we multiply $\left[\begin{array}{c}\langle \epsilon_1| \\ \vdots \\ \langle \epsilon_n| \end{array}\right]$ from the right on both side of Eq. (3) and use the completeness property (2), we obtain \begin{equation}  \cal{O}\, =\Big[|\epsilon_1\rangle\,\,|\epsilon_2\rangle\,\,\cdots\,\,|\epsilon_n\rangle\Big]\begin{bmatrix} o_{11} & o_{12} & \cdots & o_{1n} \\ o_{21} & o_{22} & \cdots & o_{2n} \\ \vdots  & \vdots  & \ddots & \vdots  \\ o_{n1} & o_{n2} & \cdots & o_{nn} \end{bmatrix}\left[\begin{array}{c}\langle \epsilon_1| \\ \langle \epsilon_2 \\ \vdots \\ \langle \epsilon_n| \end{array}\right]\,,\end{equation}that is \begin{equation} \cal{O}=\sum_{i,j=1}^n|\epsilon_i\rangle\,o_{ij}\,\langle\epsilon_j|\,.\tag{5}\end{equation}
Note that Eq. (5) is usually obtained by inserting the completeness relation (2) on both sides of the operator: \begin{equation}\cal{O}=\left(\sum_{i=1}^n|\epsilon_i\rangle\langle \epsilon_i|\right)\cal{O}\left(\sum_{j=1}^n|\epsilon_j\rangle\langle \epsilon_j|\right)=\sum_{i,j=1}^n|\epsilon_i\rangle\langle \epsilon_i|\cal{O}|\epsilon_j\rangle\langle \epsilon_j|=\sum_{i,j=1}^n|\epsilon_i\rangle o_{ij}\langle \epsilon_j|\,.\end{equation}

Operators in Quantum Mechanics

Adjoint Operator

Given an inner product, any operator $\cal{O}$ in the form of Eq. (5) defines an adjoint operator \begin{equation}\cal{O}^{\dagger}\equiv \sum_{i,j=1}^n|\epsilon_i\rangle\,o^*_{ji}\,\langle\epsilon_j|\,,\tag{6}\end{equation} so that \begin{equation} \Big(\cal{O}|\phi\rangle\,,|\psi\rangle \Big)=\Big(|\phi\rangle\,,\cal{O}^{\dagger}|\psi\rangle \Big)\,.\tag{7}\end{equation}

[Exercise 1] Verify that Eq. (7) holds.
 
[Solution] \begin{eqnarray}\text{LHS}&=&\Big(\sum_{i,j=1}^n|\epsilon_i\rangle o_{ij}\langle \epsilon_j|\phi\rangle\,,|\psi\rangle\Big)=\sum_{i,j=1}^n\phi_j^* o^*_{ij}\Big(|\epsilon_i\rangle\,,|\psi\rangle\Big)=\sum_{i,j=1}^n\phi_j^* o^*_{ij}\psi_i\,,\\ \text{RHS}&=&\Big(|\phi\rangle, \sum_{i,j=1}^n|\epsilon_i\rangle\,o^*_{ji}\,\langle\epsilon_j|\psi\rangle\Big)=\sum_{i,j=1}^n o^*_{ij}\psi_j \Big( |\phi\rangle\,, |\epsilon_i\rangle\Big)=\sum_{i,j=1}^n\phi_i^* o^*_{ji}\psi_j\,.\end{eqnarray}

Hermitian Operator

A Hermitian operator $\cal{H}$ satisfies the relation \begin{equation} \Big(\cal{H}|\phi\rangle\,,|\psi\rangle \Big)= \Big(|\phi\rangle\,,\cal{H}|\psi\rangle \Big)\,.\tag{8}\end{equation} That is, a Hermitian operator can always be flipped over to the other side in an inner product. Because of Eq. (7), an equivalent definition of Hermitian operators is \begin{equation}\cal{H}=\cal{H}^{\dagger}\,.\end{equation}

[Exercise 2] Prove that all the eigenvalues of Hermitian operators are real.

[Solution] For an eigenstate $|\lambda_i\rangle$ that satisfies  $\cal{H}|\lambda_i\rangle=\lambda_i\,|\lambda_i\rangle$, we can compute \begin{eqnarray}\Big(\cal{H}|\lambda_i\rangle\,,\, |\lambda_i\rangle\Big)&=& \Big(\lambda_i\, |\lambda_i\rangle\,,\, |\lambda_i\rangle\Big)=\lambda^*_i\,\Big(|\lambda_i\rangle,\, |\lambda_i\rangle\Big)=\lambda^*_i\,,\\ \Big(|\lambda_i\rangle\,, \,\cal{H}|\lambda_i\rangle\Big)&=& \Big( |\lambda_i\rangle\,,\,\lambda_i\,|\lambda_i\rangle\Big)=\lambda_i\,\Big(|\lambda_i\rangle,\, |\lambda_i\rangle\Big)=\lambda_i\,.\end{eqnarray} As a result, we have $\lambda_i=\lambda^*_i$ when $\cal{H}$ is Hermitian.

[Exercise 3] Prove that eigenstates of Hermitian operators with different eigenvalues are orthogonal.

[Solution] For Hermitian operator $\cal{H}$, there is relation \begin{equation} \Big(\cal{H}|\lambda_i\rangle\,,\, |\lambda_j\rangle\Big)=\Big( |\lambda_i\rangle\,,\,\cal{H}|\lambda_j\rangle\Big)\,.\end{equation} The left side is $\lambda^*_i \Big(|\lambda_i\rangle\,,\, |\lambda_j\rangle\Big)=\lambda_i\Big(|\lambda_i\rangle\,,\, |\lambda_j\rangle\Big)$ and the right side is $\lambda_j \Big(|\lambda_i\rangle\,,\, |\lambda_j\rangle\Big)$. If $\lambda_i \neq \lambda_j$, we have $\Big(|\lambda_i\rangle\,,\, |\lambda_j\rangle\Big)=0$.

Unitary Operator

An unitary operator $\cal{U}$ preserves the inner product \begin{equation}\Big(\cal{U}|\phi\rangle\,,\cal{U}|\psi\rangle \Big)= \Big(|\phi\rangle\,,|\psi\rangle \Big)\,.\tag{9}\end{equation}. Because of Eq. (7), an equivalent definition of unitary operators is \begin{equation}\cal{U}\cal{U}^{\dagger}=\cal{U}^{\dagger}\cal{U}=\cal{I}\,.\end{equation}

As shown in this post, sometimes it is easier to prove the unitarity by verifying Eq. (9).

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