**Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers **

**Required physics: Schrödinger equation, probability density **

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.9.

Suppose a particle is in the quantum state

where is the normalization constant and is a constant with dimensions of 1/time. We can find from normalization:

The spatial component of the wave function is

and it must satisfy the time-independent Schrödinger equation in one dimension

The energy can be found from the time equation:

where

Therefore

From 7 we have

This is the harmonic oscillator potential, and the wave function is actually the ground state of that potential.

We can work out a few average values:

since is even.

The standard deviations are

and the uncertainty principle is

so in this case, the uncertainty is the minimum possible.

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Igor QuinteroCould you please show how you obtained in further detail?

gwrowePost authorWhich equation are you referring to?

Igor QuinteroSorry, I think the system took my numbers and tried to convert it into an equation haha…

Specifically steps 11/12/13.

gwrowePost authorEquation 11 is the second derivative of equation 6, eqn 12 is eqn 7 solved for . You can then substitute eqns 6, 10 and 11 into 12 to get 13.

Igor QuinteroThanks!

Also, it might be a good idea to mention you used Integration by Parametric Derivatives for step 15.

gwrowePost authorActually I used the Maple software package, but if you wanted to do it by hand then you’d use integration by parts (never heard it called parametric derivatives, though, but I assume that’s what you mean).

Igor QuinteroYou can do it by integration by parts but sometimes it’s not possible (or its very difficult) in which case Integration using Parametric derivatives is better. (I find the latter a MUCH more powerful tool that can be applied without the use of computer programs, which is excellent for University students)

https://en.wikipedia.org/wiki/Integration_using_parametric_derivatives

Cindyequation 15~17 using the notation {\psi\left(x\right)} (small Psi)

did you include A in it??

or i will get the same result without considering A

i did these calculation by hand using Big Psi and i don’t feel i will get the same result if i miss A @@

thank u >W<

gwrowePost authoris given by equation 6 which includes .

MateusHello, isn’t equation (15) wrong??

Shouldn’t it be: [; = \frac{ \hbar^2 \pi}{8 m^2 a^2} ;] ??

Because [; \psi ^2 = Sqrt[\frac{\hbar \pi}{2 m a}] ;]

and [; \int x^2 \psi^2 dx = [ \frac{\hbar}{2 m a} ]^(3/2) * \frac{\pi}{2} ;]

MateusSorry for the formatation, here it goes correct:

Hello, isn’t equation (15) wrong??

Shouldn’t it be: [; = \frac{ \hbar^2 \pi}{8 m^2 a^2} ??

Because [; \psi ^2 = \sqrt{\frac{\hbar \pi}{2 m a}} exp[- \frac{2 a m}{\hbar} x^2} ) ;]

and [; \int x^2 exp[- \frac{2 a m}{\hbar} x^2} ) dx = \left[ \frac{\hbar}{2 m a} \right] ^{3/2} * \frac{\sqrt{\pi}}{2} ;]

gwrowePost authorSorry, I can’t read your Latex code properly so I’m not entirely sure what it says. However, the answer you give () cannot be correct as you can see if you check the units. must have the units of but your answer has units of . I did the integral in equation 15 using Maple, and the units check out, so I’m pretty sure it’s right.

MateusSorry for the format. You’re right, I was making a mistake with the constants. Sorry, and thank you, 🙂

Chern XiDear prof, what is the reasoning behind”the wave function is actually the ground state of that potential.” ? Should I understand it as a consequence of E=0 in equation (7)?

Also, can we further deduce that since E=0, the particle is in its ground state where its uncertainty of both momentum and position are minimum, which in turns proving that (sigma x)(sigma p) must be greater than

( h-bar /2)?

gwrowePost authoris not zero (see equation 10). The statement after equation 13 just points out that the given wave function is the ground state of the harmonic oscillator, which we worked out earlier (follow the link in the post).

Chern XiThank prof. Just now I was carelessly mess up the cancel out of h(alpha) in v(x) as E=0.

KayHow did you solve the integral in Equation 3 to get Equation 4?

gwrowePost authorI used Maple, but the integral is a Gaussian integral so it’s not too hard to do it by hand.

Bill ChristopherHow did u solve the integral of 15-17 by hand? Can u please explain me?

gwrowePost authorJust plug in from equation 6. By using integration by parts, you can reduce them to standard Gaussian integrals, which you can look up in a table.

Bill ChristopherAhh i see. Massive thank you

Joey ContrerasI think I’m missing something in regards to equation (13). When I solved equation (12) I had what you had for (13) except there was also an extra term E. So I had V(x) = 2m(a^2)(x^2) + (h-bar)(a). Did I make some mistake in my math?

gwrowePost authorIf you plug (11) and (6) into (12) you should get (13). Check your math.

sermShouldn’t the exp. val. of (equation 17) be multiplied by the normalization constant also?

(thus, = m*h_bar*a*sqrt(2*a*m/pi*h_bar))

gwrowePost authorEquation 17 is correct. Just plug in from equation 6 and most of the constants cancel out.