Normalizing the Wavefunction: A Question of Integration

In summary, the conversation discusses normalizing a wavefunction and the process of evaluating the integral to find the constant A. After some steps and corrections, the final answer is found to be A = √λ.
  • #1
says
594
12

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.
 
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  • #2
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
 
  • #3
A2 -e2λx/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...
 
  • #4
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?
 
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Likes says
  • #5
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 +∞

= - e2λx / 2λ (from -∞ to 0, x will always be negative: -x*-2λ = 2λx)

- e2λx / 2λ | from -∞ to 0
limit as x approaches -∞ = C
Now evaluating the integral from -∞ to 0

- e2λ0 / 2λ = - e0 / 2λ = -1 / 2λ

= (- e2λC / 2λ) - (-1/2λ)

= [(- e2λC / 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)

= [ -e2λC/2λ + 1/2λ ]

Now [1] + [2]


[(- e2λC / 2λ) + (1/2λ)] + [ (-e2λC/2λ) + (1/2λ) ]

I think I have made a sign error somewhere else with the - e2λC / 2λ term

I think the - e2λC / 2λ term should cancel and I should be left with A2λ=1...∴A=√(1/λ)

I hope my formatting isn't too messy
 
  • #6
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.

[tex]
\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\:
[/tex]

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

Evaluating the integrals,

[tex]

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

\\~\\

\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)

[/tex]

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.
 
  • Like
Likes says
  • #7
RedDelicious said:
Evaluating the integrals,

[tex]

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

\\~\\

\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)

[/tex]

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
 
  • #8
says said:
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.
 
  • Like
Likes says
  • #9

What is normalizing the wavefunction?

Normalizing the wavefunction refers to the process of adjusting the amplitude of the wavefunction so that its total probability is equal to 1. This allows for the accurate calculation of probabilities for different states of a quantum system.

Why is normalizing the wavefunction important?

Normalizing the wavefunction is important because it ensures that the total probability of all possible states of a quantum system adds up to 1. This is necessary for accurate predictions and calculations using quantum mechanics.

How do you normalize a wavefunction?

To normalize a wavefunction, you must first calculate its total probability by taking the square of its amplitude and integrating it over all possible states. Then, divide the wavefunction by the square root of the total probability to obtain the normalized wavefunction.

What happens if you do not normalize the wavefunction?

If the wavefunction is not normalized, it means that the total probability of all possible states does not add up to 1. This can lead to inaccurate predictions and calculations in quantum mechanics.

Can a wavefunction be normalized to a value other than 1?

No, the wavefunction must always be normalized to a value of 1. This is because the total probability of all possible states must always add up to 1 in quantum mechanics.

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