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2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is?
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2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with...
Solution:
To find the shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0), we need to consider the time evolution of the state ψ(t).
Time Evolution of a Quantum State:
The time evolution of a quantum state ψ(t) is given by the Schrödinger equation:
iħ(dψ(t)/dt) = Hψ(t)
where i is the imaginary unit, ħ is the reduced Planck's constant, and H is the Hamiltonian operator of the system.
Given Information:
We are given that ψ1 and ψ2 are the normalized eigenfunctions of a particle with energies E1 and E2 respectively, where E2 > E1. Also, at t = 0, the particle is prepared in the state ψ(t = 0) = 1/√2(ψ1 + ψ2).
Time Evolution of ψ(t = 0):
Using the time evolution operator, we can find the state ψ(t) at any time t:
ψ(t) = e^(-iHt/ħ)ψ(t = 0)
Finding the Orthogonality Condition:
To find the shortest time T at which ψ(t = T) is orthogonal to ψ(t = 0), we need to calculate the inner product of ψ(t = 0) and ψ(t = T) and set it to zero:
⟨ψ(t = 0)|ψ(t = T)⟩ = 0
Calculating the Inner Product:
Using the time evolution operator, we can calculate ψ(t = T):
ψ(t = T) = e^(-iHT/ħ)ψ(t = 0)
Now, let's calculate the inner product:
⟨ψ(t = 0)|ψ(t = T)⟩ = ⟨ψ(t = 0)|e^(-iHT/ħ)ψ(t = 0)⟩
Applying the Orthogonality Condition:
Setting the inner product to zero, we have:
⟨ψ(t = 0)|e^(-iHT/ħ)ψ(t = 0)⟩ = 0
Calculating the Exponential:
To simplify the calculation, let's express the exponential term as a power series:
e^(-iHT/ħ) = 1 - (iHT/ħ) + (1/2!)(iHT/ħ)^2 - (1/3!)(iHT/ħ)^3 + ...
Substituting this into the inner product expression, we get:
⟨ψ(t = 0)|(1 - (iHT/ħ) + (1/2!)(iHT/ħ)^2 - (1/3!)(iHT/ħ)^3 + ...)ψ(t = 0)⟩ = 0
Calculating the Inner Product:
Expanding the inner product, we have:
⟨ψ(t = 0)|ψ(t = 0)⟩ - (iHT/ħ)⟨ψ(t = 0)|ψ
To find the shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0), we need to consider the time evolution of the state ψ(t).
Time Evolution of a Quantum State:
The time evolution of a quantum state ψ(t) is given by the Schrödinger equation:
iħ(dψ(t)/dt) = Hψ(t)
where i is the imaginary unit, ħ is the reduced Planck's constant, and H is the Hamiltonian operator of the system.
Given Information:
We are given that ψ1 and ψ2 are the normalized eigenfunctions of a particle with energies E1 and E2 respectively, where E2 > E1. Also, at t = 0, the particle is prepared in the state ψ(t = 0) = 1/√2(ψ1 + ψ2).
Time Evolution of ψ(t = 0):
Using the time evolution operator, we can find the state ψ(t) at any time t:
ψ(t) = e^(-iHt/ħ)ψ(t = 0)
Finding the Orthogonality Condition:
To find the shortest time T at which ψ(t = T) is orthogonal to ψ(t = 0), we need to calculate the inner product of ψ(t = 0) and ψ(t = T) and set it to zero:
⟨ψ(t = 0)|ψ(t = T)⟩ = 0
Calculating the Inner Product:
Using the time evolution operator, we can calculate ψ(t = T):
ψ(t = T) = e^(-iHT/ħ)ψ(t = 0)
Now, let's calculate the inner product:
⟨ψ(t = 0)|ψ(t = T)⟩ = ⟨ψ(t = 0)|e^(-iHT/ħ)ψ(t = 0)⟩
Applying the Orthogonality Condition:
Setting the inner product to zero, we have:
⟨ψ(t = 0)|e^(-iHT/ħ)ψ(t = 0)⟩ = 0
Calculating the Exponential:
To simplify the calculation, let's express the exponential term as a power series:
e^(-iHT/ħ) = 1 - (iHT/ħ) + (1/2!)(iHT/ħ)^2 - (1/3!)(iHT/ħ)^3 + ...
Substituting this into the inner product expression, we get:
⟨ψ(t = 0)|(1 - (iHT/ħ) + (1/2!)(iHT/ħ)^2 - (1/3!)(iHT/ħ)^3 + ...)ψ(t = 0)⟩ = 0
Calculating the Inner Product:
Expanding the inner product, we have:
⟨ψ(t = 0)|ψ(t = 0)⟩ - (iHT/ħ)⟨ψ(t = 0)|ψ
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2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is?
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2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is?.
2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is?.
Solutions for 2. Let ψ1 and ψ2 denote the normalized eigenfunctions of particle with energy E1 and E2 respectively, with E2¿E1. At time t=0 the particle is prepared in the state ψ(t = 0) = 1 2 (ψ1 ψ2). The shortest time T at which ψ(t = T) will be orthogonal to ψ(t = 0) is? in English & in Hindi are available as part of our courses for GATE.
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