Probability Calculation Using Orthonormal Basis

The state of the quantum system is given by the state vector:

$|\psi\rangle = |\phi_1\rangle+2|\phi_2\rangle+3|\phi_3\rangle$

The basis states $|\phi_1\rangle$, $|\phi_2\rangle$, and $|\phi_3\rangle$ form an orthonormal set. This means their inner products satisfy $\langle \phi_i | \phi_j \rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta ($\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i \neq j$).

Determining Probability

The probability, $P_i$, of finding the system in a specific state $|\phi_i\rangle$ when it is in state $|\psi\rangle$ is calculated using the formula:

$P_i = \frac{|\langle \phi_i | \psi \rangle|^2}{\langle \psi | \psi \rangle}$

First, calculate the inner product of the target state $|\phi_2\rangle$ with the system state $|\psi\rangle$:

$\langle \phi_2 | \psi \rangle = \langle \phi_2 | (|\phi_1\rangle+2|\phi_2\rangle+3|\phi_3\rangle)$

Using the linearity of the inner product and the orthonormality condition:

$\langle \phi_2 | \psi \rangle = \langle \phi_2 | \phi_1 \rangle + 2\langle \phi_2 | \phi_2 \rangle + 3\langle \phi_2 | \phi_3 \rangle = 0 + 2(1) + 3(0) = 2$

Next, calculate the normalization factor $\langle \psi | \psi \rangle$. For an orthonormal basis, this is the sum of the squared magnitudes of the coefficients:

$\langle \psi | \psi \rangle = |1|^2 + |2|^2 + |3|^2 = 1 + 4 + 9 = 14$

Calculating Final Probability

Now, substitute these values into the probability formula for the state $|\phi_2\rangle$ ($P_2$):

$P_2 = \frac{|\langle \phi_2 | \psi \rangle|^2}{\langle \psi | \psi \rangle} = \frac{|2|^2}{14} = \frac{4}{14} = \frac{2}{7}$

To express this as a decimal rounded to two places:

$P_2 = \frac{2}{7} \approx 0.2857...$

Rounding to two decimal places gives $0.29$.