Probability of being in a state is given, Find the normalised wavefunction

[dipanshum]

Homework Statement

A particle can be in two different states given by orthonormal wavefunctions ψ1 and ψ2. If the probability of being in state ψ1 is 1/3, find out normalized wave function for the particle.

Relevant Equations

ρ(r,t) = |ψ(r,t)|^2
where ρ= probability

Should I treat ψ₁ as ψ and ψ ₂ as ψ^*?

 

[Orodruin]

What you have written down is not the probability of finding the particle in a particular state, it is the probability density of finding the particle at position $r$. That the particle is in a state where it is in either $\psi_1$ or $\psi_2$ means that it is a linear combination of those states, i.e., $\psi = \alpha \psi_1 + \beta \psi_2$. You need to find $\alpha$ and $\beta$. (Note that there is an inherent phase ambiguity.)

 

[dipanshum]

What you have written down is not the probability of finding the particle in a particular state, it is the probability density of finding the particle at position $r$. That the particle is in a state where it is in either $\psi_1$ or $\psi_2$ means that it is a linear combination of those states, i.e., $\psi = \alpha \psi_1 + \beta \psi_2$. You need to find $\alpha$ and $\beta$. (Note that there is an inherent phase ambiguity.)

Ok, I worked out the sum. Please tell me if I'm right.
The normalized wavefunction is:
ψ = 1/3 ψ₁ + 2√2/3 ψ₂

 

[Orodruin]

How do you compute the probability of being in the state $\psi_1$?

 

[dipanshum]

How do you compute the probability of being in the state $\psi_1$?

The probability of ψ₁ is given in the question.

 

[PeroK]

The probability of ψ₁ is given in the question.

Can you check that your $\psi$ has this probability? You said:

The normalized wavefunction is:
ψ = 1/3 ψ₁ + 2√2/3 ψ₂

What are the respective probabilities of finding the system in $\psi_1$ and $\psi_2$ using your $\psi$?

 

[Orodruin]

The probability of ψ₁ is given in the question.

Yes, but your job is to show that your state satisfies that and in order to do that you must compute the probability of your particle being in the state $\psi_1$.

 

[dipanshum]

Can you check that your $\psi$ has this probability? You said:
What are the respective probabilities of finding the system in $\psi_1$ and $\psi_2$ using your $\psi$?

no only probability of ψ is given, but that is sufficient data. and what you suggested already helped me and the answer is right.
what happened is after normalization one will get α²+β²=1
then calculate and put values of α, β in the equation ψ=αψ₁+βψ₂

 

[PeroK]

no only probability of ψ is given, but that is sufficient data. and what you suggested already helped me and the answer is right.
what happened is after normalization one will get α²+β²=1
then calculate and put values of α, β in the equation ψ=αψ₁+βψ₂

What's the probability of the state being $\psi_2$?

 

[Orodruin]

and the answer is right.

It is not.

 

[dipanshum]

It is not.

Okay so, yes Orodruin you are right, and I was wrong.
The final normalised function would be ψ=1/√3 ψ₁ + √(2/3)ψ ₂.
Tell me if I'm right this time.