Let |\psi\rangle=\frac{1-i}{2}|0\rangle-\frac{1+i}{2}|1\rangle and |\phi\rangle=\frac{2+i}{3}|0\rangle-\frac{2i}{3}|1\rangle. What is \langle \psi\,|\,\phi\rangle?
Please tell the ans...
To find the inner product or the dot product of two vectors, we need to take the complex conjugate of one of the vectors and then multiply them together element-wise, and finally sum up the results.
Given:
|\psi\rangle=\frac{1-i}{2}|0\rangle-\frac{1+i}{2}|1\rangle
|\phi\rangle=\frac{2+i}{3}|0\rangle-\frac{2i}{3}|1\rangle
First, let's take the complex conjugate of |\psi\rangle:
\langle \psi | = \left( \frac{1+i}{2} \right)^* \langle 0 | + \left( \frac{1-i}{2} \right)^* \langle 1 |
= \frac{1-i}{2^*} \langle 0 | + \frac{1+i}{2^*} \langle 1 |
= \frac{1+i}{2} \langle 0 | + \frac{1-i}{2} \langle 1 |
Next, we can multiply |\phi\rangle by \langle \psi | element-wise and sum up the results:
\langle \psi | \phi \rangle = \left( \frac{1+i}{2} \langle 0 | + \frac{1-i}{2} \langle 1 | \right) \left( \frac{2+i}{3}|0\rangle-\frac{2i}{3}|1\rangle \right)
= \frac{1+i}{2} \cdot \frac{2+i}{3} \langle 0 | 0 \rangle + \frac{1+i}{2} \cdot \left( -\frac{2i}{3} \right) \langle 0 | 1 \rangle + \frac{1-i}{2} \cdot \frac{2+i}{3} \langle 1 | 0 \rangle + \frac{1-i}{2} \cdot \left( -\frac{2i}{3} \right) \langle 1 | 1 \rangle
Now, let's evaluate each term:
\langle 0 | 0 \rangle = 1, since it represents the inner product of the basis vector |0\rangle with itself, which is equal to 1.
\langle 0 | 1 \rangle = 0, since it represents the inner product of different basis vectors |0\rangle and |1\rangle, which is zero.
\langle 1 | 0 \rangle = 0, same as above.
\langle 1 | 1 \rangle = 1, since it represents the inner product of the basis vector |1\rangle with itself, which is equal to 1.
Now, plugging in these values:
\langle \psi | \phi \rangle = \frac{1+i}{2} \cdot \frac{2+i}{3} \cdot 1 + \frac{1+i}{2} \cdot \left( -\frac{2i}{3} \right) \cdot 0 + \frac{1-i}{2} \cdot \frac{2+i}{3} \cdot 0 + \frac{1-i}{2} \cdot \left( -\frac{2i}{3} \right) \cdot 1
Simplifying this expression:
\langle \psi | \phi \rangle = \frac{1+i}{2} \cdot \frac{2+i}{3} - \frac{1-i}{2} \cdot \frac{2i}{3}
Multiplying the complex numbers:
\langle \psi | \phi \rangle = \frac{1+i}{2} \cdot \frac{2}{3} - \frac{1+i}{2} \cdot \frac{i}{3} - \frac{1-i}{2} \cdot \frac{2i}{3} - \frac{1-i}{2} \cdot \frac{-i}{3}
= \frac{2+2i}{6} - \frac{i-1}{6} + \frac{2i-2}{6} - \frac{-i-1}{6}
= \frac{2+2i - i + 1 + 2i - 2 + i + 1}{6}
= \frac{6i}{6}
= i
Therefore, the value of \langle \psi | \phi \rangle is i.