Integration help with normalizing wave function

[leroyjenkens]

Homework Statement

ψ(x,t) = Ae^-λxe^-iwt

Normalize this and solve for A

Homework Equations

[tex]\int_{-∞}^{∞}|ψ|²dx = 1<h2>The Attempt at a Solution</h2><br /> I got to A²$\int_{-∞}^{∞}e^{-2λx}dx$<br /> <br /> The solution manual multiplies it by 2 and only goes from 0 to ∞ instead of from -∞ to ∞. I'm trying to do it from -∞ to ∞ and I should get the same answer as the manual, but I'm not. They get a finite non-zero answer, but integrating what I got to, I have $-\frac{A^{2}}{2λ}e^{-2λx}$ and now, plugging in the limits of integration in for x, it will give me infinity as the answer.<br /> Anyone see where I made a mistake? Thanks.[/tex]

 

[jbunniii]

I got to A²$\int_{-∞}^{∞}e^{-2λx}dx$

The solution manual multiplies it by 2 and only goes from 0 to ∞ instead of from -∞ to ∞. I'm trying to do it from -∞ to ∞ and I should get the same answer as the manual, but I'm not. They get a finite non-zero answer, but integrating what I got to, I have $-\frac{A^{2}}{2λ}e^{-2λx}$ and now, plugging in the limits of integration in for x, it will give me infinity as the answer.
Anyone see where I made a mistake? Thanks.

The integral as written above is infinite. It is not the same as integrating from $0$ to $\infty$ and multiplying by $2$. Just plot the integrand to see this. If the integrand were $e^{-2\lambda |x|}$ (with absolute values) then it would be true.

 

[leroyjenkens]

The integral as written above is infinite. It is not the same as integrating from $0$ to $\infty$ and multiplying by $2$. Just plot the integrand to see this. If the integrand were $e^{-2\lambda |x|}$ (with absolute values) then it would be true.

It is supposed to be absolute value. I keep forgetting about that. But now I'm getting 0 as the answer, since the ∞ in the exponent will make it look like this when I evaluate the integral over the limits:

$$-\frac{A^{2}}{2λ}[e^{-2λ∞}-e^{-2λ∞}]$$

Which is just

$$-\frac{A^{2}}{2λ}[0 - 0]$$

Which equals 0.

I know I'm doing something wrong, though.

Thanks for the response.

 

[jbunniii]

It is supposed to be absolute value. I keep forgetting about that. But now I'm getting 0 as the answer, since the ∞ in the exponent will make it look like this when I evaluate the integral over the limits:

$$-\frac{A^{2}}{2λ}[e^{-2λ∞}-e^{-2λ∞}]$$

What integral did you perform? What were the upper and lower limits?

 

[leroyjenkens]

What integral did you perform? What were the upper and lower limits?

Upper limit was ∞ and lower was -∞. I plug them in for x, and since it's the absolute value of x, they're both just ∞, which means it turns to 0.

 

[jbunniii]

Upper limit was ∞ and lower was -∞. I plug them in for x, and since it's the absolute value of x, they're both just ∞, which means it turns to 0.

So you are trying to perform this integral:
$$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx$$
Because of the absolute value, you can't evaluate the integral without breaking it into parts. (If you believe otherwise, then you need to produce an antiderivative for $e^{-2\lambda|x|}$.) Try writing it this way:
$$\int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx =
\int_{-\infty}^{0} e^{-2 \lambda |x|} dx + \int_{0}^{\infty} e^{-2 \lambda |x|} dx$$
Next step: simplify the integrand in each case by replacing $|x|$ with something simpler.