Griffiths 1.5: Normalization of a wave function

[Radarithm]

Homework Statement

Consider the wave function $$\Psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}$$
Where $A$, $\lambda$, and $\omega$ are positive real constants.

(a)Normalize $\Psi$.
(b)Determine expectation values of $x$ and $x^2$.

Homework Equations

$$\int_{-\infty}^{+\infty}|\Psi(x,t)|^2dx=\int_{-\infty}^{+\infty}\Psi\Psi^{\ast}dx=1$$

The Attempt at a Solution

Correct me if I'm wrong, but in order to normalize the wave function you need to re-write A in terms of the other constants and $x$, and normalize it at t = 0. This is what I did:

$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx$$
$$|A|^2\int_{-\infty}^{+\infty}e^{-\lambda^2|x|^2}dx$$
$$|A|^2\frac{\sqrt{\pi}}{2\lambda}=1$$
$$A=\frac{\sqrt{2\lambda}}{(\pi^{1/4})}$$

Did I go wrong somewhere in (a) ?

 

[micromass]

Homework Statement

Consider the wave function $$\Psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}$$
Where $A$, $\lambda$, and $\omega$ are positive real constants.

(a)Normalize $\Psi$.
(b)Determine expectation values of $x$ and $x^2$.

Homework Equations

$$\int_{-\infty}^{+\infty}|\Psi(x,t)|^2dx=\int_{-\infty}^{+\infty}\Psi\Psi^{\ast}dx=1$$

The Attempt at a Solution

Correct me if I'm wrong, but in order to normalize the wave function you need to re-write A in terms of the other constants and $x$, and normalize it at t = 0. This is what I did:

$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx$$
$$|A|^2\int_{-\infty}^{+\infty}e^{-\lambda^2|x|^2}dx$$

In this step, did you do

$$(e^x)^2 = e^{x^2}$$

because that would be incorrect.

 

[Radarithm]

In this step, did you do

$$(e^x)^2 = e^{x^2}$$

because that would be incorrect.

I did, I should have payed more attention to the math;

Correcting my mistake:
$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx=|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx$$
$$|A|^2=\frac{e^{-2\lambda|x|}}{-2\lambda}$$
From negative infinity to positive infinity.
$$|A|^2=\frac{1}{2\lambda}$$
$$|A|=\sqrt{\frac{1}{2\lambda}}$$

Is this correct? Or did I just formally prove the fact that I suck at integration?

 

[Saitama]

I did, I should have payed more attention to the math;

Correcting my mistake:
$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx=|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx$$
$$|A|^2=\frac{e^{-2\lambda|x|}}{-2\lambda}$$

I do not see how you get that. Notice that:

$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx$$
Can you integrate this?

 

[micromass]

I did, I should have payed more attention to the math;

Correcting my mistake:
$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx=|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx$$
$$|A|^2=\frac{e^{-2\lambda|x|}}{-2\lambda}$$

This cannot be correct, the formula for $|A|$ can not contain an $x$.

You need to integrate

$$\int_0^{+\infty} e^{-2\lambda x}dx + \int_{-\infty}^0 e^{2\lambda x}dx$$

 

[Radarithm]

This cannot be correct, the formula for $|A|$ can not contain an $x$.

You need to integrate

$$\int_0^{+\infty} e^{-2\lambda x}dx + \int_{-\infty}^0 e^{2\lambda x}dx$$

That was what I did, and I got:
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$
and that is:
$$-\infty-(-\frac{1}{2\lambda})+\frac{1}{2\lambda}-\infty$$
Which means that
$$A=\sqrt{\frac{1}{\lambda}}$$
because the wave function disappears at plus or minus infinity.

 

[micromass]

That was what I did, and I got:
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$
and that is:
$$-\infty-(-\frac{1}{2\lambda})+\frac{1}{2\lambda}-\infty$$

Not a fan of that notation. You shouldn't use infinite that way.

You know the integral you need to find is
$$\int e^{-2\lambda x}dx = -\frac{e^{2\lambda x}}{2\lambda} + C$$

So indeed, the integral I want you to find is
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$

But this is equal to

$$\lim_{x\rightarrow +\infty} -\frac{e^{-2\lambda x}}{2\lambda} + \frac{e^{-2\lambda 0}}{2\lambda}+\frac{e^{2\lambda 0}}{2\lambda} - \lim_{x\rightarrow -\infty}\frac{e^{2\lambda x}}{2\lambda}$$

Both limits are $0$ if $\lambda>0$. Thus the above expression is $1/\lambda$

Which means that
$$A=\sqrt{\frac{1}{\lambda}}$$
because the wave function disappears at plus or minus infinity.

You have now calculated

$$\int_{-\infty}^{+\infty} |\Psi(x,0)|^2 dx = |A|^2 \int_{-\infty}^{+\infty} e^{-2\lambda |x|}dx = \frac{|A|^2}{\lambda}$$

You want this to be equal to $1$, so what should $|A|$ be?

 

[Radarithm]

I do not see how you get that. Notice that:

$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx$$
Can you integrate this?

Now I see where I was wrong:
$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx=2|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=0-(-\frac{1}{2\lambda})$$
So:
$$|A|=\sqrt{\lambda}$$

 

[micromass]

Now I see where I was wrong:
$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx=2|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=0-(-\frac{1}{2\lambda})$$
So:
$$|A|=\sqrt{\lambda}$$

Right, this is the right result.

 

[Radarithm]

Not a fan of that notation. You shouldn't use infinite that way.

You know the integral you need to find is
$$\int e^{-2\lambda x}dx = -\frac{e^{2\lambda x}}{2\lambda} + C$$

So indeed, the integral I want you to find is
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$

But this is equal to

$$\lim_{x\rightarrow +\infty} -\frac{e^{-2\lambda x}}{2\lambda} + \frac{e^{-2\lambda 0}}{2\lambda}+\frac{e^{2\lambda 0}}{2\lambda} - \lim_{x\rightarrow -\infty}\frac{e^{2\lambda x}}{2\lambda}$$

Both limits are $0$ if $\lambda>0$. Thus the above expression is $1/\lambda$



You have now calculated

$$\int_{-\infty}^{+\infty} |\Psi(x,0)|^2 dx = |A|^2 \int_{-\infty}^{+\infty} e^{-2\lambda |x|}dx = \frac{|A|^2}{\lambda}$$

You want this to be equal to $1$, so what should $|A|$ be?

That was the dumbest mistake I made this year. I set A equal to 1 over lambda when I should have multiplied them and set them equal to one. Thanks for the help, though.

 

[micromass]

So, what about the expectations?

 

[Radarithm]

So, what about the expectations?

Working them out now..

 

[Radarithm]

When solving for expectation values, is t set to zero as well? Is it so it is possible to integrate with respect to x?

 

[micromass]

When solving for expectation values, is t set to zero as well? Is it so it is possible to integrate with respect to x?

It doesn't matter whether you set $t$ equal to $0$ here. You need to take the modulus of $\Psi(x,t)$ anyway, and

$$|\Psi(x,t)| = |A| e^{-\lambda|x|} |e^{i\omega t}| = |A| e^{-\lambda |x|}$$

So the terms with $t$ drop anyway.

 

[Radarithm]

$$\langle x\rangle=\int_{-\infty}^{+\infty}x\lambda e^{-2\lambda|x|}dx$$
Thus:
$$\langle x \rangle=\left(-\frac{e^{-2\lambda x}}{4\lambda}|_0^{+\infty}\right)+\left(\frac{e^{2\lambda x}}{4\lambda}|_{-\infty}^0\right)$$
$$\langle x\rangle=\frac{1}{2\lambda}$$

Right?

 

[micromass]

$$\langle x\rangle=\int_{-\infty}^{+\infty}x\lambda e^{-2\lambda|x|}dx$$
Thus:
$$\langle x \rangle=\left(-\frac{e^{-2\lambda x}}{4\lambda}|_0^{+\infty}\right)+\left(\frac{e^{2\lambda x}}{4\lambda}|_{-\infty}^0\right)$$
$$\langle x\rangle=\frac{1}{2\lambda}$$

Right?

The function $x\lambda e^{-2\lambda |x|}$ is odd. So what does that mean about the integral? I think you must have made a sign error somewhere. You should get

$$\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda}$$

and

$$\int_{-\infty}^0 x\lambda e^{-2\lambda x}dx = -\frac{1}{4\lambda}$$

 

[Radarithm]

The function $x\lambda e^{-2\lambda |x|}$ is odd. So what does that mean about the integral? I think you must have made a sign error somewhere. You should get

$$\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda}$$

and

$$\int_{-\infty}^0 x\lambda e^{-2\lambda x}dx = -\frac{1}{4\lambda}$$

I see my mistake. Also, how does the function being odd or even affect the integral? I'm a self learner so I'm not surprised at the fact that my knowledge is fragmented.
I'm guessing this means that the expectation value is 0. How can this be?

 

[micromass]

I see my mistake. Also, how does the function being odd or even affect the integral? I'm a self learner so I'm not surprised at the fact that my knowledge is fragmented.
I'm guessing this means that the expectation value is 0. How can this be?

It's a very easy test to compute integrals. It's always worth checking this before doing a long computation.

So, if a function is odd (meaning that $f(-x) = -f(x)$), then

$$\int_{-d}^d f(x)dx = 0$$

for all $d$ including $d=+\infty$. This can be seen by splitting up the integral into

$$\int_{-d}^0 f(x)dx + \int_0^d f(x)dx$$

and by applying substitution $x\rightarrow -x$ on the first integral.

You can also check the graph to see this easily.

If a function is even, then we can't compute the integral this way, but we can make some simplification:

$$\int_{-d}^d f(x)dx = 2\int_0^d f(x)dx$$

The proof is the same as in the case of odd functions.

 

[Saitama]

If a function is even, then we can't compute the integral this way, but we can make some simplification:

$$\int_{-d}^d f(x)dx = 2\int_0^d f(x)dx$$

The proof is the same as in the case of odd functions.

@Radarithm: This is what I used when I wrote the definite integral. Can you relate?

 

[Radarithm]

@Radarithm: This is what I used when I wrote the definite integral. Can you relate?

Apparently that was an even function. Changing the sign of $x$ would not affect the entire function.

 

[Saitama]

Apparently that was an even function.

You still need to calculate the expectation for $x^2$. :)

 

[micromass]

You still need to calculate the expectation for $x^2$. :)

Check first whether that will be an even or odd function (or neither)

 

[Radarithm]

You still need to calculate the expectation for $x^2$. :)

Working on that. Looks like I'm going to use an integral calculator or look it up in a table; it seems quite messy.

 

[Radarithm]

Check first whether that will be an even or odd function (or neither)

It's odd.

 

[Saitama]

Working on that. Looks like I'm going to use an integral calculator or look it up in a table; it seems quite messy.

No, it isn't, it should be doable using integration by parts. What integral do you get?

 

[micromass]

It's odd.

It's not odd, it's even

 

[Radarithm]

It's not odd, it's even

You're right, I confused the wave function with the expectation.

 

[Radarithm]

No, it isn't, it should be doable using integration by parts. What integral do you get?

Integrating by parts twice (first using $x^2\lambda=u$ and the second term as dv), I get:
$$\frac{x^2e^{-2\lambda|x|}}{2}+\frac{xe^{-2\lambda|x|}}{2\lambda}-\frac{e^{-2\lambda|x|}}{4\lambda^2}$$

 

[Saitama]

Integrating by parts twice (first using $x^2\lambda=u$ and the second term as dv), I get:
$$\frac{x^2e^{-2\lambda|x|}}{2}+\frac{xe^{-2\lambda|x|}}{2\lambda}-\frac{e^{-2\lambda|x|}}{4\lambda^2}$$

You don't actually need to evaluate the antiderivative. You can use the result micromass posted in his post #16.

The integral you have is:
$$\int_{-\infty}^{\infty} x^2\lambda e^{-2\lambda |x|}\,dx=2\int_0^{\infty} x^2\lambda e^{-2\lambda x}\,dx$$
From post #16:
$$\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda} \Rightarrow \int_0^{+\infty} xe^{-2\lambda x}dx = \frac{1}{4\lambda^2}$$
Differentiate both the sides with respect to $\lambda$ to obtain:
$$\int_0^{\infty} -2x^2 e^{-2\lambda x}\,dx=\frac{-1}{2\lambda^3}\Rightarrow \int_0^{\infty} x^2\lambda e^{-2\lambda x}\,dx=\frac{1}{4\lambda^2} $$
From here, you can get the result.

You can still evaluate it using integration by parts though, I just wanted to show a quicker method.

 

[Radarithm]

$$\langle x^2\rangle=\frac{1}{2\lambda^2}$$