Normalizing the Wavefunction: A Question of Integration

[says]

Homework Statement

Consider the wavefunction:

Ψ(x, t) = Ae^-λ|x|e^−iωt

Normalise Ψ(x, t)

Homework Equations

∫ |Ψ(x, t)|² dx = 1 (bounds from x: -∞ to +∞)

The Attempt at a Solution

Ψ(x, t) = Ae^-λ|x|e^−iωt

Ψ(x, t) = Ae^{-λ|x|−iωt}
Ψ*(x, t) = Ae^-λ|x|+iωt

∫ |Ψ(x, t)|² dx = A² ∫ e^{-λ|x|−iωt+(-λ|x|+iωt)} dx

= A² ∫ e^{-λ|x|−iωt-λ|x|+iωt} dx

= A² ∫ e^-λ|x|-λ|x| dx

= A² ∫ e^-2λ|x| dx

Ok, so evaluating this integral (bounds from x: -∞ to +∞)

= A² (-e^-2λx/2λ) | x: -∞ to +∞

I tried to evaluate the integral from -∞ to 0 + 0 to +∞, but I got A²*0 = 1. I've gone through the problem a few times and I'm not really sure where I went wrong. I'm trying to do this without the help of a calculator as well. Any help would be much appreciated.

 

[RedDelicious]

You can't just ignore the absolute value function. The absolute value function is a piecewise function, meaning you will have to split up and evaluate your integral accordingly.

Note:

|x| = x for x>0 and -x for x<0

 

[says]

A² -e^2λx/2λ (from -∞ to 0, x will always be minus: -x*-2λ = 2λx)

A² -e^-2λx/2λ (from 0 to +∞, x will always be positive)

I still get the same result...

 

[RedDelicious]

You're making a simple computational error somewhere then. Do you mind showing your full evaluation of the integral so I can see what you're doing wrong?

 

[says]

Ψ(x, t) = Ae^-λ|x|e^−iωt

Ψ(x, t) = Ae^{-λ|x|−iωt}
Ψ*(x, t) = Ae^-λ|x|+iωt

∫ |Ψ(x, t)|² dx = A² ∫ e^{-λ|x|−iωt+(-λ|x|+iωt)} dx

= A² ∫ e^{-λ|x|−iωt-λ|x|+iωt} dx

= A² ∫ e^-λ|x|-λ|x| dx

Ok, so the integral now. I'm ignoring the A² for the moment just to make things a bit simpler

∫ e^-2λ|x| dx

= (- e^-2λx / 2λ) | x: -∞ to +∞

Change the bounds so its -∞ to 0 and 0 to +∞

= - e^2λx / 2λ (from -∞ to 0, x will always be negative: -x*-2λ = 2λx)

- e^2λx / 2λ | from -∞ to 0
limit as x approaches -∞ = C
Now evaluating the integral from -∞ to 0

- e^2λ0 / 2λ = - e⁰ / 2λ = -1 / 2λ

= (- e^2λC / 2λ) - (-1/2λ)

= [(- e^2λC / 2λ) + (1/2λ)] [integral 1]

---------------------------

-e^-2λx/2λ (from 0 to +∞, x will always be positive)(I think this may have been where I made an error)

= -e^-2λx/2λ | from 0 to ∞
limit as x approaches ∞ = C
Now evaluating the integral from 0 to ∞

=(-e^-2λC/2λ) - (-e⁰/2λ)

=(-e^-2λC/2λ) - (-e⁰/2λ)

=(-e^-2λC/2λ) - (-1/2λ)

= [ -e^2λC/2λ + 1/2λ ]

Now [1] + [2]


[(- e^2λC / 2λ) + (1/2λ)] + [ (-e^2λC/2λ) + (1/2λ) ]

I think I have made a sign error somewhere else with the - e^2λC / 2λ term

I think the - e^2λC / 2λ term should cancel and I should be left with A²λ=1...∴A=√(1/λ)

I hope my formatting isn't too messy

 

[RedDelicious]

Ok, so a few things here.

Firstly, just as a note, you can normalize the wave function at t=0 and it'll be normalized for all t, meaning you can immediately simplify the integral some and avoid some of the extra algebra.

Now as for the calculation, your evaluation for the piecewise integral is still wrong.

<br /> \int _{-\infty }^{\infty }\:e^{-2\lambda \left|x\right|}dx=\int _{-\infty }^0e^{2\lambda x}dx+\int _0^{\infty }e^{-2\lambda x}dx\:<br />

What you're doing is taking the antiderivative of a different function and then just matching the bounds piecewise afterward.

Evaluating the integrals,

<br /> <br /> \int _{-\infty }^0e^{2\lambda x}dx=\frac{1}{2\lambda }\lim _{t\to -\infty }\left(e^0-e^t\right)<br /> <br /> \\~\\<br /> <br /> \int _0^{\infty }e^{-2\lambda x}dx=\frac{-1}{2\lambda }\lim _{t\to \infty }\left(e^{-t}-e^0\right)=\frac{1}{2\lambda }\lim _{t\to \infty }\left(e^0-e^{-t}\right)<br /> <br />

Also, realize that in both cases of the limit you get the exponential tending to negative infinity. One is -x tending to infinity and the other is x tending to negative infinity, which are of course, identical. What is the value of this limit? Take a look a the graph of the exponential. At the end, I dropped the factor in the exponential because it doesn't matter in limit.

 

[says]

Evaluating the integrals,

<br /> <br /> \int _{-\infty }^0e^{2\lambda x}dx=\frac{1}{2\lambda }\lim _{t\to -\infty }\left(e^0-e^t\right)<br /> <br /> \\~\\<br /> <br /> \int _0^{\infty }e^{-2\lambda x}dx=\frac{-1}{2\lambda }\lim _{t\to \infty }\left(e^{-t}-e^0\right)=\frac{1}{2\lambda }\lim _{t\to \infty }\left(e^0-e^{-t}\right)<br /> <br />

I think I can see where I made a mistake now. I didn't change the signs for the two different integrals. I don't really understand the notation (above) though.

I thought the limit as x approaches negative and positive infinity = 0.

A²*[1/2λ + 1/2λ] = A²*[1/λ]

A = √λ

 

[RedDelicious]

I think I can see where I made a mistake now. I didn't change the signs for the two different integrals. I don't really understand the notation (above) though.

I thought the limit as x approaches negative and positive infinity = 0.

A²*[1/2λ + 1/2λ] = A²*[1/λ]

A = √λ

That is the correct answer. Both the exponentials go to zero as they approach negative infinity. I only left them in the limit above as to make sure that you knew that.

 

[says]

Thanks for your help @RedDelicious