Correct Answer - Option 1 :
\(\frac{1}{\sqrt{5}}\)
Concept:
- we interpret \(|\psi(x)|^2 = \psi(x)^* \psi(x)\)as defining a probability distribution for finding the particle at some position x.
- since particle should exist somewhere, so if we sum up the probabilities over all the position of x they should sum to 1 (100% probability)
-
Normalizing a wave function means multiplying it by a constant, to ensure that the sum of the probabilities for finding the particle should be equal to 1
- mathematically it is written as \(\int_{-\infty}^{\infty} \psi(x)^* \psi(x) = 1\)
- In Bra-Ket notation it is written as \(<\psi(x)|\psi(x)> = 1\)also known as the inner product of wave function
-
Orthogonal: when we multiply two vectors which is perpendicular to each other the dot product turn out to be Zero
- similarly, in quantum machines, the inner product of wave-function which are in the different state has a value of zero
-
Mathematically can be represented as \(<\psi_n|\psi_m> = 0 \\ where\: m \neq n\)
- If \(<\psi_n|\psi_m> = 0 \: where\: m \neq n, \\ <\psi_n|\psi_n> = 1 \)then they are called orthonormal.
- The wave function is orthonormal to each other
Explanation:
- Given:\( \psi(x) = A[\phi_1(x) + 2\phi_2(x)] \)
- To find: Normalization constant
the normalization condition in Bar-Ket notation is, \(<\psi(x)|\psi(x)> = 1\)
\(1= A^2<\phi_1 + 2\phi_2|\phi_1+ 2\phi_2>\) taking inner product
\(1=A^2 <\phi_1|\phi_1> + <2\phi_2|\phi_1> + <\phi_1|2\phi_2> + <2\phi_2|2\phi_2>\)
By applying orthogonality condition we get,
\(1= A^2 [1 + 4] \)...................................\(<\psi_n|\psi_m> = 0 \: where\: m \neq n, \\ <\psi_n|\psi_n> = 1 \)
Thus \(A = \frac{1}{\sqrt{5}}\)
hence option 1 is correct
