How Is the Ground State Wave Function for a 1-D SHM Oscillator Determined?

[noblegas]

Homework Statement

the ground state wave function for 1-d SHM oscillator is fixed by the diff. eqn a*phi_-0=0using the expression for the lower operator as a differential operator,a_-=(K/2)^1/2x-h*($$\partial$$)/(2pi*(2m)^1/2), to find a solution for this differential equation for phi₀(x)

integral(-infinity,infinity)e^-y2=pi^1/2 to normalize phi₀ to the condition integral(-infinity,infinity)abs(phi₀)²dx=1

Homework Equations

The Attempt at a Solution

a_-*phi₀=0 =>( K/2)^1/2x-h*($$\partial$$)/(2pi*(2m)^1/2)[/SUB]phi₀=0

I don't understand how the three equations a_-*phi₀=0, integral(-infinity,infinity)abs(phi₀)²dx=1,and -infinity,infinity)e^-y2=pi^1/2 are related;

 

[gabbagabbahey]

Homework Statement

the ground state wave function for 1-d SHM oscillator is fixed by the diff. eqn a*phi_-0=0


using the expression for the lower operator as a differential operator,a_-=(K/2)^1/2x-h*($$\partial$$)/(2pi*(2m)^1/2), to find a solution for this differential equation for phi₀(x)

integral(-infinity,infinity)e^-y2=pi^1/2 to normalize phi₀ to the condition integral(-infinity,infinity)abs(phi₀)²dx=1

Homework Equations

The Attempt at a Solution

a_-*phi₀=0 =>( K/2)^1/2x-h*($$\partial$$)/(2pi*(2m)^1/2)[/SUB]phi₀=0

I don't understand how the three equations a_-*phi₀=0, integral(-infinity,infinity)abs(phi₀)²dx=1,and -infinity,infinity)e^-y2=pi^1/2 are related;

Yuck!

I think it will help you to better communicate future homework problems if you format your posts a little nicer. That includes using $\LaTeX$ when appropriate, and making an effort to use proper spelling and grammar.

Here's a nice introduction to using $\LaTeX$ in these forums.

For example, your post could have looked something like this:

Homework Statement

The ground state wave function for 1-D SHM oscillator is fixed by the differential equation $a_{-}|\varphi_0\rangle=0$


Use the expression for the lowering operator as a differential operator (in the x-basis),

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}$$

to find a solution for this differential equation for $\varphi_0(x)$

Use $\int_{-\infty}^{\infty}e^{-y^2}=\pi^{1/2}$ to normalize $\varphi_0(x)$ to the condition $\int_{-\infty}^{\infty} |\varphi_0(x)|^2 dx=1$

Homework Equations

$$a_{-}|\varphi_0\rangle=0$$

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}$$

$$\int_{-\infty}^{\infty}e^{-y^2}dy=\pi^{1/2}$$

$$\int_{-\infty}^{\infty} |\varphi_0(x)|^2 dx=1$$

The Attempt at a Solution

$$a_{-}|\varphi_0\rangle=0 \implies \left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0$$

I don't understand how the other three equations in the 'relevant equations' section are related

You can click on any of the above images to see the code that generated them.

Now, as for your question... You have a differential equation,

$$\left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0$$

Just solve it!...You will find that the solution has an unknown constant in front of it...you can then use the normalization condition to find the value of that constant.

 

[noblegas]

Yuck!

I think it will help you to better communicate future homework problems if you format your posts a little nicer. That includes using $\LaTeX$ when appropriate, and making an effort to use proper spelling and grammar.

Here's a nice introduction to using $\LaTeX$ in these forums.

For example, your post could have looked something like this:



You can click on any of the above images to see the code that generated them.

Now, as for your question... You have a differential equation,

$$\left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0$$

Just solve it!...You will find that the solution has an unknown constant in front of it...you can then use the normalization condition to find the value of that constant.

What am I solving for? \varphi_0(x)? And if I am solving for \varphi_0(x) would I then plug that into my normalization equation to obtain k

 

[gabbagabbahey]

What am I solving for? \varphi_0(x)?

Yes. Your problem statement says to 'find a solution for $\varphi_0(x)$'

And if I am solving for \varphi_0(x) would I then plug that into my normalization equation to obtain k

Not quite...try solving for $\varphi_0(x)$ first, worry about the rest later.

 

[noblegas]

Yes. Your problem statement says to 'find a solution for $\varphi_0(x)$'



Not quite...try solving for $\varphi_0(x)$ first, worry about the rest later.

I think I eventually found a solution to for $\varphi_0(x)$; I won't list my math; $\varphi_0(x)$=A*exp(-(m)^(1/2)*(k)^(1/2)/h); I would used , $$\int_{-\infty}^{\infty} |\varphi_0(x)|^2 dx=1$$

to find A?

 

[gabbagabbahey]

I think I eventually found a solution to for $\varphi_0(x)$; I won't list my math; $\varphi_0(x)$=A*exp(-(m)^(1/2)*(k)^(1/2)/h)

No, why is there an 'h' and an 'm' in your solution? And why is there no 'x'?

Make sure you are using the correct expression for the differential operator $a_{-}$ in the x-basis.

This can either be written as

$$a_{-}\to\left(\frac{m\omega}{2\hbar}\right)^{1/2}x-\left(\frac{\hbar}{2m\omega}\right)^{1/2}\frac{d}{d x}$$

or

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}$$

with the constant $k$ defined by $k\equiv\frac{m\omega}{\hbar}$

I would used ,$\int_{-\infty}^{\infty} |\varphi_0(x)|^2 dx=1$ to find A?

Once you find the correct expression for $\varphi_0(x)$, yes.

 

[noblegas]

No, why is there an 'h' and an 'm' in your solution? And why is there no 'x'?

Make sure you are using the correct expression for the differential operator $a_{-}$ in the x-basis.

This can either be written as

$$a_{-}\to\left(\frac{m\omega}{2\hbar}\right)^{1/2}x-\left(\frac{\hbar}{2m\omega}\right)^{1/2}\frac{d}{d x}$$

or

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}$$

with the constant $k$ defined by $k\equiv\frac{m\omega}{\hbar}$
Once you find the correct expression for $\varphi_0(x)$, yes.

Would this expression $$\int_{-\infty}^{\infty}e^{-y^2}dy=\pi^{1/2}$$ be useful and helping me find A? Wouldn't I squared this expression and pi^(1/2) becomes Pi?

 

[gabbagabbahey]

You still haven't correctly found $\varphi_0(x)$...do that first, worry about normalizing it later.

 

[noblegas]

You still haven't correctly found $\varphi_0(x)$...do that first, worry about normalizing it later.

I got $\varphi_0(x)$ to be: $\varphi_0(x)$=A*exp(-(x^2/2)*(K/2)^(1/2)*((2m)^(1/2)/$$\hbar$$)

 

[gabbagabbahey]

I got $\varphi_0(x)$ to be: $\varphi_0(x)$=A*exp(-(x^2/2)*(K/2)^(1/2)*((2m)^(1/2)/$$\hbar$$)

That's closer, but your constants inside the exponential are incorrect. You should be getting
$$\varphi_0(x)=Ae^{-kx^2/2}$$

Or, in terms of $\omega$,

$$\varphi_0(x)=Ae^{-\frac{m\omega x^2}{2\hbar}}$$

Are you still using the incorrect expression

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{\hbar}{2m}\right)^{1/2}\frac{d}{d x}$$

?

If so, refer to post #6 for the correct expressions, or better yet, look them up in your textbook.

 

[noblegas]

$$a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{\hbar}{2m}\right)^{1/2}\frac{d}{d x}$$?

If so, refer to post #6 for the correct expressions, or better yet, look them up in your textbook.

that is the expression given in the textbook except the Planck constant is not supposed raised by the 1/2 power

 

[gabbagabbahey]

That's odd, which textbook are you using?

 

[noblegas]

That's odd, which textbook are you using?

peebles

 

[gabbagabbahey]

I found a google book review of that text, which page is this equation on in your copy?

 

[noblegas]

I found a google book review of that text, which page is this equation on in your copy?

p.37, eqn 6.34

 

[gabbagabbahey]

Okay, I see what he's doing now, he is using a rather odd definition of the raising and lowering operators which results in $\omega=\sqrt{\frac{K}{m}}$

In that case, your earlier result,

$$\varphi_0=Ae^{-\frac{\sqrt{mK}x^2}{2\hbar}}$$

is correct.(I simplified it a little for you)

Using this result, what do you get when you evaluate the integral $\int_{-\infty}^{\infty}|\varphi_0(x)|^2 dx$ ?

 

[noblegas]

Okay, I see what he's doing now, he is using a rather odd definition of the raising and lowering operators which results in $\omega=\sqrt{\frac{K}{m}}$

In that case, your earlier result,

$$\varphi_0=Ae^{-\frac{\sqrt{mK}x^2}{2\hbar}}$$

is correct.(I simplified it a little for you)

Using this result, what do you get when you evaluate the integral $\int_{-\infty}^{\infty}|\varphi_0(x)|^2 dx$ ?

$$\varphi_0^2$$dx=$$A^2e^{-\frac{{mK}x^4}{4\hbar^2}}$$dx =1?

Sorry about latex; I been fumbling with it to a point where I want to put latex to rest for good with a dagger

 

[gabbagabbahey]

No, first off

$$(e^u)^2=e^{2u}\neq e^{u^2}$$

Secondly, you need to actually perform the integration/

 

[noblegas]

No, first off

$$(e^u)^2=e^{2u}\neq e^{u^2}$$

Secondly, you need to actually perform the integration/

I should have known better; would I normaliased this equation;
$\int_{-\infty}^{\infty}e^{-y^2}=\pi^{1/2}$ and y^2 =2u?

 

[gabbagabbahey]

First off, what is $|\varphi_0(x)^2|$?

Second, what do you get when you integrate it from -infinity to infinity?

 

[noblegas]

First off, what is $|\varphi_0(x)^2|$?

Second, what do you get when you integrate it from -infinity to infinity?

I think $|\varphi_0(x)^2|$ is the probability function; obviously, when I integrate the probability function from -infinity to infinity I get 1 as stated in the problem;

 

[queenofbabes]

Yup, you get 1=) Do you have any problems with the integration itself?

 

[noblegas]

Yup, you get 1=) Do you have any problems with the integration itself?

yes; I think I am supposed to square this expression
$\int_{-\infty}^{\infty}e^{-y^2}=\pi^{1/2}$ and plugged 2u into y .

 

[gabbagabbahey]

yes; I think I am supposed to square this expression
$\int_{-\infty}^{\infty}e^{-y^2}=\pi^{1/2}$ and plugged 2u into y .

Why on Earth would you do that?

You have found an expression for $\varphi_0$,

$$\varphi_0=Ae^{-\frac{\sqrt{mK}x^2}{2\hbar}}$$

You now need to normalize that expression.

That means that you need to actually integrate $\int_{-\infty}^{\infty}|\varphi_0(x)|^2 dx$ using your expression for $\varphi_0$, and then determine the value of $A$ by setting the result equal to one...so go integrate!