Griffiths 1.5: Normalization of a wave function

In summary: I learned something.In summary, the conversation discusses normalizing a wave function with the given equation and determining expectation values for x and x^2. The individual makes a mistake in calculating the value for A in part (a), but is corrected by another individual.
  • #1
Radarithm
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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) ?
 
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  • #2
Radarithm said:

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

[tex](e^x)^2 = e^{x^2}[/tex]

because that would be incorrect.
 
  • #3
micromass said:
In this step, did you do

[tex](e^x)^2 = e^{x^2}[/tex]

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?
 
  • #4
Radarithm said:
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?
 
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  • #5
Radarithm said:
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

[tex]\int_0^{+\infty} e^{-2\lambda x}dx + \int_{-\infty}^0 e^{2\lambda x}dx[/tex]
 
  • #6
micromass said:
This cannot be correct, the formula for ##|A|## can not contain an ##x##.

You need to integrate

[tex]\int_0^{+\infty} e^{-2\lambda x}dx + \int_{-\infty}^0 e^{2\lambda x}dx[/tex]

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.
 
  • #7
Radarithm said:
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
[tex]\int e^{-2\lambda x}dx = -\frac{e^{2\lambda x}}{2\lambda} + C[/tex]

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

[tex]\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}[/tex]

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

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

You want this to be equal to ##1##, so what should ##|A|## be?
 
  • #8
Pranav-Arora said:
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:
$$|A|=\sqrt{\lambda}$$
 
  • #9
Radarithm said:
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.
 
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  • #10
micromass said:
Not a fan of that notation. You shouldn't use infinite that way.

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

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

[tex]\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}[/tex]

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



You have now calculated

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

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.
 
  • #11
So, what about the expectations?
 
  • #12
micromass said:
So, what about the expectations?

Working them out now..
 
  • #13
When solving for expectation values, is t set to zero as well? Is it so it is possible to integrate with respect to x?
 
  • #14
Radarithm said:
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

[tex]|\Psi(x,t)| = |A| e^{-\lambda|x|} |e^{i\omega t}| = |A| e^{-\lambda |x|}[/tex]

So the terms with ##t## drop anyway.
 
  • #15
$$\langle x\rangle=\int_{-\infty}^{+\infty}x\lambda e^{-2\lambda|x|}dx$$
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?
 
  • #16
Radarithm said:
$$\langle x\rangle=\int_{-\infty}^{+\infty}x\lambda e^{-2\lambda|x|}dx$$
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

[tex]\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda}[/tex]

and

[tex]\int_{-\infty}^0 x\lambda e^{-2\lambda x}dx = -\frac{1}{4\lambda}[/tex]
 
  • #17
micromass said:
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

[tex]\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda}[/tex]

and

[tex]\int_{-\infty}^0 x\lambda e^{-2\lambda x}dx = -\frac{1}{4\lambda}[/tex]

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?
 
  • #18
Radarithm said:
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

[tex]\int_{-d}^d f(x)dx = 0[/tex]

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

[tex]\int_{-d}^0 f(x)dx + \int_0^d f(x)dx[/tex]

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:

[tex]\int_{-d}^d f(x)dx = 2\int_0^d f(x)dx[/tex]

The proof is the same as in the case of odd functions.
 
  • #19
micromass said:
If a function is even, then we can't compute the integral this way, but we can make some simplification:

[tex]\int_{-d}^d f(x)dx = 2\int_0^d f(x)dx[/tex]

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?
 
  • #20
Pranav-Arora said:
@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.
 
  • #21
Radarithm said:
Apparently that was an even function.

You still need to calculate the expectation for ##x^2##. :)
 
  • #22
Pranav-Arora said:
You still need to calculate the expectation for ##x^2##. :)

Check first whether that will be an even or odd function (or neither)
 
  • #23
Pranav-Arora said:
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.
 
  • #24
micromass said:
Check first whether that will be an even or odd function (or neither)

It's odd.
 
  • #25
Radarithm said:
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?
 
  • #26
Radarithm said:
It's odd.

It's not odd, it's even
 
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  • #27
micromass said:
It's not odd, it's even

You're right, I confused the wave function with the expectation.
 
  • #28
Pranav-Arora said:
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:
$$\frac{x^2e^{-2\lambda|x|}}{2}+\frac{xe^{-2\lambda|x|}}{2\lambda}-\frac{e^{-2\lambda|x|}}{4\lambda^2}$$
 
  • #29
Radarithm said:
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.
 
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  • #30
$$\langle x^2\rangle=\frac{1}{2\lambda^2}$$
:approve:
 
  • #31
Radarithm said:
When solving for expectation values, is t set to zero as well?
In general no, you will have to keep the t dependence. (Your system may not be in a stationary state of the Hamiltonian for example.)
Also, in your example, if you had to compute say, ##\langle p \rangle## then by differentiating once, you pull down another factor before the exponents cancel.
 
  • #32
Can I just inject a little physics into this discussion. I see a lot of math, and that's all great, but some physical intuition would be something good to have. Let's examine the given wave function.

1) First notice its time dependence. The time dependence of this wave function is merely an overall phase factor. There can be no room for interference effects since the same phase factor multiplies the entire expression. Therefore, we know that this state is, in fact, a stationary state wherein all the expectation values of all operators will be time independent. So, looking for the expectation values, I should expect no dependence on t. This is NOT true for general states. A general state may have expectation values dependent on t, so that you may have to find something like ##\left<x(t)\right>##.

2) Notice that the wave function is an even function of x. In other words, x is symmetric about x=0. This tells you that I have as much chance of finding the particle to the left of x=0 as to the right of x=0. From this consideration alone, without doing ANY calculations, I can immediately conclude ##\left<x\right>=0##.

3) Notice that the wave function is "spread out" (i.e. it's not a delta function at some x). Therefore, we can conclude that the statistical properties of this wave function are such that if I measured x multiple times on identical copies of this wave function, that I should get some different values of x. The standard deviation of my statistical ensemble of measurement should be ##\sigma_x=\sqrt{\left<x^2\right>-\left<x\right>^2}\neq 0## . Thus, from this argument alone, I can see that ##\left<x^2\right>\neq 0##.
 
  • #33
Matterwave said:
2) Notice that the wave function is an even function of x. In other words, x is symmetric about x=0. This tells you that I have as much chance of finding the particle to the left of x=0 as to the right of x=0. From this consideration alone, without doing ANY calculations, I can immediately conclude ##\left<x\right>=0##.

I'm going to be nitpicky and not agree with this. For example the Cauchy distribution http://en.wikipedia.org/wiki/Cauchy_distribution is symmetric around ##0##. But the mean does not exist, since the integral does not exist.
 
  • #34
micromass said:
I'm going to be nitpicky and not agree with this. For example the Cauchy distribution http://en.wikipedia.org/wiki/Cauchy_distribution is symmetric around ##0##. But the mean does not exist, since the integral does not exist.

Well, you could use Cauchy's principal value to redefine that integral in which case the mean turns out to be zero as expected. At any rate, this is quite a pathological distribution. The main use of that distribution is to showcase the possibility of pathological distributions...
 

1. What is a wave function?

A wave function is a mathematical representation of a quantum mechanical system that describes the probability of finding a particle at a specific location and time.

2. Why is normalization important for a wave function?

Normalization ensures that the total probability of finding a particle in all possible locations is equal to 1, as required by the laws of quantum mechanics.

3. How do you normalize a wave function?

To normalize a wave function, you need to calculate its integral over all space and then divide the wave function by the square root of the integral. This ensures that the total probability is equal to 1.

4. What is the significance of the normalization constant in a wave function?

The normalization constant ensures that the wave function is properly normalized and represents a valid probability distribution. It also affects the overall shape and behavior of the wave function.

5. Can a wave function be normalized to a value other than 1?

No, the total probability of finding a particle must always be equal to 1, so the wave function must be normalized to this value. Normalizing to a different value would violate the principles of quantum mechanics.

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