How do I normalize Ψ(x,0) for a particle in an infinite square well?

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Your Name]In summary, the conversation discusses normalising a wave function Ψ(x,0) in an infinite square well of width a, using the fact that ψ1 and ψ2 are stationary states. The approach involves integrating the square of the wave function between 0 and a and equating it to 1. The question also mentions the confusion about integrating ψ1(x) or ψ2(x), which can be treated as constants since they are stationary states. The conversation also addresses a question about the cancellation of the exponential term when squaring the complex conjugate, which will not cancel out in the integration but will cancel out when integrating over time.
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maxbye3
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Homework Statement


Normalise Ψ(x,0) (use the fact that both ψ1 and ψ2 are stationary states).
Will this wave function be normalised at any later time, t > 0?

Homework Equations


A particle in an infinite square well of width a has as its initial wave function an even mixture of the first two stationary states,
Ψ(x, 0) = A(ψ1(x) + ψ2(x)) .

The Attempt at a Solution


I get the concept of normalisation I would integrate the square of A(ψ1(x) + ψ2(x)) between 0 and a and equate it to 1.
I am confused how to integrate ψ1(x) or ψ2(x).
I see that ψ(n)=(2/L)^0.5 sin(π/L)exp−i(n2π2 ̄h)/2mL2)t
does the exponential term cancel when I square the complex conjugate?Anyway any help would be deeply appreciated, thanks.
 
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Hello,
Yes, you are correct in your approach to normalising the wave function. To integrate ψ1(x) or ψ2(x), you can use the fact that they are both stationary states, meaning they do not change with time. This allows you to treat them as constants when integrating.
As for your question about the exponential term, when you take the complex conjugate and square it, the exponential term will become positive, so it will not cancel out. However, when you integrate over time, the exponential term will still cancel out because it will have a factor of e^(-iEn t/ħ) and e^(iEn t/ħ), which will cancel each other out.
I hope this helps. Let me know if you have any other questions.

 

FAQ: How do I normalize Ψ(x,0) for a particle in an infinite square well?

What is the process for integrating ψ1(x) / ψ2(x)?

The process for integrating ψ1(x) / ψ2(x) involves finding the antiderivative of the quotient, which is also known as the indefinite integral. This can be done by using integration techniques such as u-substitution or integration by parts.

How do I know when to use u-substitution or integration by parts?

The decision to use u-substitution or integration by parts depends on the specific form of the quotient ψ1(x) / ψ2(x). Generally, if the quotient contains a single function raised to a power, u-substitution is a good technique to use. If the quotient contains a product of two functions, integration by parts is a better choice.

Is there a general formula for integrating ψ1(x) / ψ2(x)?

No, there is not a general formula for integrating ψ1(x) / ψ2(x). The integration process will vary depending on the specific form of the quotient and may require the use of different integration techniques.

Can I use a calculator to integrate ψ1(x) / ψ2(x)?

Yes, there are many online calculators that can help with the integration process. However, it is important to have a basic understanding of integration techniques and be able to verify the results from the calculator.

Are there any common mistakes to avoid when integrating ψ1(x) / ψ2(x)?

Some common mistakes when integrating ψ1(x) / ψ2(x) include forgetting to add the constant of integration, making errors in the algebraic manipulation of the integrand, and not checking the final result for accuracy. It is important to double check each step of the integration process to avoid these mistakes.

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