How to Tackle Quantum Mechanics Integrals for Modern Physics Exam?

[Shackleford]

URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

I've done reasonably well in the class so far. However, I have not done well since starting modern quantum mechanics. The last several problem sets have covered quantum mechanics, and I failed Problem Set 6. I've never failed anything in college. The last exam in Modern Physics I (no comprehensive final) is tomorrow. I know this thread is a bit lengthy, and I'd be willing to paypal some $$$ if someone wants to really help me out. If I'm having this much trouble in the beginning of modern quantum mechanics, I've deeply worried about Modern Physics II next semester.

Let's just focus on the integration, particularly the wave function in momentum space integrals.

PS6.2 - I'm not sure how she got lines two and three.

PS6.5 - I have no clue what she did.

PS6.8 - I setup the initial integral correctly, but I don't know how she got the two integrals.

PS6.16 - Again, I setup the initial integral correctly, but I'm not sure how she got the final integral.

PS7.5 - I don't know what I did wrongly in my computation.

PS7.6 - Again, I don't know what I did wrongly in my computation.

PS7.14 - No clue on how she did the integral. This is clearly a recurring theme for me. I'm not sure how to do these integrals.

Problem Set 6: #2, 5, 8, 16

Problems

My Pitiful Work

Solutions Posted by TA

Problem Set 7: #5, 6, 14

Problems

My Pitiful Work

Solutions Posted by TA

 

[zachzach]

Line 2 comes directly from the question. All she did was square both sides of the given expression and and solve for $$v^2 - {v_0}^2$$.

For the third line:

$$v_g = -{\lambda}^2\frac{dv}{d\lambda}$$

$$v^2 - {v_0}^2 = \frac{c^2}{{\lambda}^2}$$

Take the derivative:

$$2v\frac{dv}{d\lambda} - 0 = \frac{-2c^2}{{\lambda}^3} \rightarrow \frac{dv}{d\lambda} = -\frac{c^2}{v{\lambda}^3}$$

Plug that into

$$v_g = -{\lambda}^2\frac{dv}{d\lambda}$$

And then plug the original equation for $$\lambda$$ into it.

 

[Shackleford]

Zach, thanks. I understand everything except the derivative. Why is the zero there?

I'm a bit disappointed I couldn't even pull the first correction line out of my butt.

 

[zachzach]

Zach, thanks. I understand everything except the derivative. Why is the zero there?

Because $$v_0$$ is just a constant. I tried to figure some of the other problems out but failed miserably .

 

[Shackleford]

Because $$v_0$$ is just a constant. I tried to figure some of the other problems out but failed miserably .

Ah. Of course it's a constant.

Do those momentum-space wave function or expectation-value integrals make sense to you?

 

[zachzach]

Ah. Of course it's a constant.

Do those momentum-space wave function integrals make sense to you?

Well the reason you split them up is because is because the absolute value. Just like if
$$f(x) = |x|$$

If you integrated that from -2 to 2 you would have to split it up into two different integrals. When x < 0 you have to integrate y = -x and when x > 0 you have to integrate y = x. so
$$<br /> \int_{-2}^{2} |x| dx = \int_{-2}^{0} -x dx + \int_{0}^{2} x dx$$I'm not sure if this is what you were asking though.

 

[Shackleford]

Ah. Because it's the absolute value of x?

 

[zachzach]

Ah. Because it's the absolute value of x?

Yes. Look of the graph of |x| compared to just x. It might help you visualize it. Also graph $$e^{-x}$$ and $$e^{-|x|}$$ and see the difference. Especially in integrating the function.

 

[Shackleford]

Yes. Look of the graph of |x| compared to just x. It might help you visualize it. Also graph $$e^{-x}$$ and $$e^{-|x|}$$ and see the difference. Especially in integrating the function.

Of course, they give very different function values. Dammit. I forget these important details and it completely screws me up.

Now, what about #14 and 16? I think #5 might be a lost cause for me. lol.

 

[zachzach]

Of course, they give very different function values. Dammit. I forget these important details and it completely screws me up.

Now, what about #14 and 16? I think #5 might be a lost cause for me. lol.

I am terrible at Quantum lol. But on 16 why did you change the limits from -inf -> inf into
0 -> inf?

Also do you get the odd function thing?

I must go i'll try to solve over the night.

 

[Shackleford]

I am terrible at Quantum lol. But on 16 why did you change the limits from -inf -> inf into
0 -> inf?

Also do you get the odd function thing?

What's your major/degree?

I shouldn't have changed the integration limits. I did that on a few ones. lol.

Yeah, I saw that in the book. If you integrate an odd function over symmetric limits, you get a zero value. I assume the oddness of the function is based on the exponent of x. x^1 and x^17 are odd, but x^2 is even.

 

[BerryBoy]

For question 14, I think you're correct just need the trick for the Fourier transform of a Gaussian...

So you have the Fourier transform on your wavepacket:

$$\mathcal{F}\left[ \psi(x) \right]= \left( \frac{\alpha}{\pi}\right)^{\frac{1}{4}}\frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} \exp\left[-\frac{\alpha}{2}x^2 + i \frac{p}{\hbar}x \right]\,\rm{d}x$$

We can re-write the exponential term by using a complete the squares method ie..

$$-\frac{\alpha}{2}x^2 + i \frac{p}{h}x = -\frac{\alpha}{2}\left(x - \frac{ip}{\hbar\alpha} \right)^2 + \frac{p^2}{\hbar^2 \alpha^2}$$

And the other crucial point for evaluating this integral is

$$\sqrt{\frac{\alpha}{2\pi}} \int^\infty_{-\infty} \exp\left[ -\frac{\alpha}{2}(x - \frac{ip}{\hbar\alpha})^2\right] \,\rm{d}x = 1$$

Just noticed that my integral slightly differs from your TA solution, the trick is fine just check whether I missed a symbol or not.

Hope this helps, and good luck

 

[BerryBoy]

For question 16, I started with

$$\left\langle x^n \right\rangle = \int^{\infty}_{-\infty}\psi^\ast(x) x^n \psi(x) \rm{d}x$$

So that immediately we notice that this function is anti-symmetric when n is odd. In such case, for every point +x there is a point of equal and opposite value at -x. Thus the integral over all space = 0

For n=2 this function is symmetric and so you can evaluate the integral as twice the integral from $x= 0$ to $x = +\infty$, which is a well-defined standard integral. I can't decipher how the TA solution got there, but I get the same answer.

 

[Feldoh]

What's your major/degree?

I shouldn't have changed the integration limits. I did that on a few ones. lol.

Yeah, I saw that in the book. If you integrate an odd function over symmetric limits, you get a zero value. I assume the oddness of the function is based on the exponent of x. x^1 and x^17 are odd, but x^2 is even.

That's correct. Odd functions over even intervals (- some value to + some value) will give 0. Try integrating x on [-1,1] -- it's the same concept. That is also really nifty for evaluating higher dimension integrals.

The idea will come in handy when dealing with different quantum states.

 

[Shackleford]

Thanks for the help, guys. I'll have to look it over when I have some free time here at work.

 

[Shackleford]

For question 14, I think you're correct just need the trick for the Fourier transform of a Gaussian...

So you have the Fourier transform on your wavepacket:

$$\mathcal{F}\left[ \psi(x) \right]= \left( \frac{\alpha}{\pi}\right)^{\frac{1}{4}}\frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} \exp\left[-\frac{\alpha}{2}x^2 + i \frac{p}{\hbar}x \right]\,\rm{d}x$$

We can re-write the exponential term by using a complete the squares method ie..

$$-\frac{\alpha}{2}x^2 + i \frac{p}{h}x = -\frac{\alpha}{2}\left(x - \frac{ip}{\hbar\alpha} \right)^2 + \frac{p^2}{\hbar^2 \alpha^2}$$

And the other crucial point for evaluating this integral is

$$\sqrt{\frac{\alpha}{2\pi}} \int^\infty_{-\infty} \exp\left[ -\frac{\alpha}{2}(x - \frac{ip}{\hbar\alpha})^2\right] \,\rm{d}x = 1$$

Just noticed that my integral slightly differs from your TA solution, the trick is fine just check whether I missed a symbol or not.

Hope this helps, and good luck

Wow. I never would have figured out what square to complete! The second term is simply a constant. Why is the first term integral equal to one? I'm not familiar with the trick of the Fourier transform of a Gaussian. It might be in m QM book, but I'll have to look.

 

[BerryBoy]

You can observe the why the first integral is equal to one by a transform of variables, from x to say y
$$y = x - \frac{ip}{\hbar \alpha}$$
$$\rm{d}y = \rm{d}x$$

So it the solution to the integral is just a standard integral with the argument that the complex term does not change your limits of integration.

$$\sqrt{\frac{\alpha}{2\pi}} \int^\infty_{-\infty}\exp\left[ -\frac{\alpha}{2}y^2\right] \rm{\, d}y = \sqrt{\frac{\alpha}{2\pi}} \sqrt{\frac{2\pi}{\alpha}}$$

Gaussian distributions are very important in quantum mechanics, maths and optics because they are Fourier transforms of themselves (in other words, your answer should certainly have the following form)

$$\psi(p) = A\exp\left[-\beta p^2\right]$$

where $A$ and $\beta$ are some constants.

 

[BerryBoy]

Wow. I never would have figured out what square to complete! The second term is simply a constant. Why is the first term integral equal to one? I'm not familiar with the trick of the Fourier transform of a Gaussian. It might be in m QM book, but I'll have to look.

Sorry, bad communication on my part. The "trick" is to re-write the equation by "completing the square" in the integrand exponential; just so that you get something you can integrate easily.