Exercise 2.5#

In the textbook, we define an operation as

(1)#\[ ((y_1, \dotsc, y_n), (z_1, \dotsc, z_n)) \equiv \sum_{i}y_i^{*}z_i \]

To check whether eq. (1) defines an inner product, we need to check the following properties of inner product.

  • \((\cdot, \cdot)\) is linear in the second argument. Suppose we have \(N\) vectors and the \(k-\)th vector is defined by

    (2)#\[ \mathbf{z}_{k} = (z_{k1}, z_{k2}, \dotsc, z_{kn}) \]

    Then we can compute

    (3)#\[ \sum_{k=1}^{N} \lambda_k\mathbf{z}_{k} = \left(\sum_{k=1}^{N}\lambda_kz_{k1}, \sum_{k=1}^{N}\lambda_kz_{k2}, \dotsc, \sum_{k=1}^{N}\lambda_kz_{kn}\right) \]

    From eq. (1), we can compute

    (4)#\[ \left((y_1, \dotsc, y_n), \sum_{k} \lambda_k\mathbf{z}_{k}\right) = \sum_{i=1}^n y_i^{*}\sum_{k=1}^{N}\lambda_kz_{k1} = \sum_{i=1}^n\sum_{k=1}^{N} y_i^{*}\lambda_kz_{k1} \]

    Meanwhile, we can compute

    (5)#\[ \sum_{k=1}^{N}\lambda_k\left((y_1, \dotsc, y_n), (z_{k1}, z_{k2}, \dotsc, z_{kn})\right) = \sum_{k=1}^{N}\lambda_k \sum_{i=1}^{n}y_i^{*}z_i = \sum_{i=1}^n\sum_{k=1}^{N} y_i^{*}\lambda_kz_{k1} \]

    From eq. (4) and eq. (5), we conclude that the given definition of \((\cdot, \cdot)\) is linear in the second argument.

  • \((|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*\). From eq. (1)​ we can compute

    (6)#\[ ((z_1, \dotsc, z_n), (y_1, \dotsc, y_n)) = \sum_{i}z_i^{*}y_i = \sum_{i}(y_i^*z_i)^* = \left(\sum_{i}y_i^*z_i\right)^* \]

    where we use relation \(\left(\sum_{i} z_i\right)^* = \sum_i z^*_i\) and \((z_1z_2)^* = z^*_1 z^*_2\) (see appendix). Compare with eq. (1) we could see that

    (7)#\[ ((z_1, \dotsc, z_n), (y_1, \dotsc, y_n)) = \left(\sum_{i}y_i^*z_i\right)^* =((y_1, \dotsc, y_n), (z_1, \dotsc, z_n))^* \]

    Therefore, we could conclude that under the definition in eq. (1), we have \((|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*\)

  • \((|v\rangle, |v\rangle) \geq 0\), and \((|v\rangle, |v\rangle) = 0\) iff \(|v\rangle = 0\). For \((|v\rangle, |v\rangle)\) we have

    (8)#\[ ((y_1, \dotsc, y_n), (y_1, \dotsc, y_n)) = \sum_{i}y_i^{*}y_i =\sum_{i} |y_i|^2 \]

    Since for any complex number \(|y_i|^2 \geq 0\), we should have \(((y_1, \dotsc, y_n), (y_1, \dotsc, y_n)) \geq 0\). Moreover,

    • If \(((y_1, \dotsc, y_n), (y_1, \dotsc, y_n)) = 0\), from eq. (8) we know that can only have \(|y_1| = \dotsc = |y_n| = 0\), or equivalently, \(|y\rangle = 0\)

    • If \(|y_1| = \dotsc = |y_n| = 0\), from eq. (8) we know that \(((y_1, \dotsc, y_n), (y_1, \dotsc, y_n)) = 0\).

    Therefore, we can conclude that with eq. (1), we have \((|v\rangle, |v\rangle) \geq 0\), and \((|v\rangle, |v\rangle) = 0\) iff \(|v\rangle = 0\)

From the proof above, we can conclude that eq. (1) defines an inner product on \(\mathbb{C}^n\).

Appendix#

We use the following two relations above to prove \((|v\rangle, |w\rangle) = (|w\rangle, |v\rangle)^*\),

\[ \left(\sum_{j} z_j\right)^* = \sum_j z^*_j\text{ and } \left(\prod_i z_i \right)^* = \prod_i z_i^* \]

To prove these two relations, we need to write complex number in the form \(z_j = x_j + iy_j\). Then we can compute

\[ \sum_{j} z_j = \sum_{j} (x_j + iy_j) = \sum_{j} x_j + i\sum_{j}y_j \]

and also,

\[ \sum_{j} z^*_j = \sum_{j} (x_j - iy_j) = \sum_{j} x_j - i\sum_{j}y_j \]

Compare above two equations, we have

\[ \left(\sum_{j} z_j \right)^* = \left(\sum_{j} x_j + i\sum_{j}y_j\right)^* = \sum_{j} x_j - i\sum_{j}y_j = \sum_j z^*_j \]

To prove the second relation, we need to firstly prove \((z_1z_2)^* = z^*_1 z_2^*\). For \((z_1z_2)^*\) we have

\[ (z_1z_2)^* = [(x_1+iy_1)(x_2+iy_2)]^* = [x_1x_2 +ix_2y_1+ix_1y_2 - y_1y_2)]^* = x_1x_2 - y_1y_2-i(x_2y_1+x_1y_2) \]

Then we have,

\[ z^*_1 z_2^* = (x_1-iy_1)(x_2-iy_2) = x_1x_2 -ix_2y_1-ix_1y_2-y_1y_2 = (z_1z_2)^* \]

For \(z^*_1 z_2^*z_3^*\) and \((z_1z_2z_3)^*\), we can make use of the relation \((z_1z_2)^* = z^*_1 z_2^*\) to prove. That is,

\[ (z_1z_2z_3)^* = (z_1z_2)^*z_3^* = z_1^*z_2^*z_3^* \]

Similarly, we can prove by induction that if

\[ \left(\prod_{i=1}^n z_i \right)^* = \prod_{i=1}^n z_i^* \]

We should have

\[ \left(\prod_{i=1}^{n+1} z_i \right)^* = \left[\left(\prod_{i=1}^{n} z_i\right) \cdot z_{n+1}\right]^* =\left(\prod_{i=1}^{n} z_i\right)^* z^*_{n+1} = \left(\prod_{i=1}^n z_i^*\right)\cdot z^*_{n+1} = \prod_{i=1}^{n+1} z_i^* \]