Help with finding the expectation value of x^2

[Lasse Jepsen]

The question is as follows:

A particle of mass m has the wave function

psi(x, t) = A * e^( -a ( ( m*x^2 / hbar) +i*t ) )

where A and a are positive real constants.

i don't know how to format my stuff on this website, so it may be a bit harder to read. Generally when i write "int" i mean the integral from -infinity to +infinity.

first it asks me to find A, i do this by normalizing the wave function, and i find that

A = (2*a*m / (pi * hbar) )^( 1 / 4)

this is correct according to the solutions manual.

i'm then asked to find the potential energy function, but since this doesn't mean anything for the rest of the question i will ignore it.

i am then asked to calculate the expectation values of x, x^2, p and p^2

so starting with the expectation value of x (i will write this as <x>) :

from calculating A i already know that int( e^( -2*a ( (m*x^2 / hbar) ) dx) = ( pi * hbar / (2*a*m) )^( 1 / 2) = A^( -2 )

<x> = int( x * |psi(x, t)|^2 dx) = A^2 int( x * e^( -2*a ( (m*x^2 / hbar) ) dx)

now i use integration by parts

int( u dv) = u*v - int( v du)

i use the following u and v:

u = x

du = dx

dv = e^( -2*a ( (m*x^2 / hbar) ) dx

v = int( dv ) = int( e^( -2*a ( (m*x^2 / hbar) ) dx) = A^( -2 )

so my integral becomes:

int( x * e^( -2*a ( (m*x^2 / hbar) ) dx) = x * A^( -2 ) - int( A^( -2 ) dx) = x * A^( -2 ) - A^( -2 ) * x = 0 = <x>

this means <x> = A^2 * 0 = 0

the solutions manual agrees with me on this solution.

but calculating <x^2> doesn't give the the correct answer, and this makes me question whether i use integration by parts wrong.

i'll show you my result for <x^2>

<x^2> = int( x^2 * |psi(x, t)|^2 dx) = A^2 int( x^2 * e^( -2*a ( (m*x^2 / hbar) ) dx)

now i use integration by parts

int( u dv) = u*v - int( v du)

i use the following u and v:

u = x^2

du = 2 * x dx

dv = e^( -2*a ( (m*x^2 / hbar) ) dx

v = int( dv ) = int( e^( -2*a ( (m*x^2 / hbar) ) dx) = A^( -2 )

so my integral becomes:

int( x^2 * e^( -2*a ( (m*x^2 / hbar) ) dx) = x^2 * A^( -2 ) - int( A^( -2 ) * 2 * x dx) = x^2 * A^( -2 ) - A^( -2 ) * 2 * int( x dx)

= x^2 * A^( -2 ) - A^( -2 ) * 2 * 1/2 * x^2 = 0 = <x^2>

but according to the solutions manual i should get

<x^2> = hbar / (4*a*m)

no matter what i do i can't seem to figure out what I'm doing wrong please help.

for anyone interested this is problem 1.9 from David J. Griffiths "introduction to Quantum Mechanics Third edition"

 

[DrClaude]

i don't know how to format my stuff on this website, so it may be a bit harder to read. Generally when i write "int" i mean the integral from -infinity to +infinity.

See https://www.physicsforums.com/help/latexhelp/

int( x^2 * e^( -2*a ( (m*x^2 / hbar) ) dx) = x^2 * A^( -2 ) - int( A^( -2 ) * 2 * x dx) = x^2 * A^( -2 ) - A^( -2 ) * 2 * int( x dx)

= x^2 * A^( -2 ) - A^( -2 ) * 2 * 1/2 * x^2 = 0 = <x^2>

There are definite integrals, so your last equality doesn't hold (you don't get a function of x).

You have to look up how to solve a Gaussian integral.

 

[mjc123]

use the following u and v:
u = x^2
du = 2 * x dx
dv = e^( -2*a ( (m*x^2 / hbar) ) dx
v = int( dv ) = int( e^( -2*a ( (m*x^2 / hbar) ) dx) = A^( -2 )

That is not legitimate, because v must be an indefinite integral, and there isn't an analytical form for the indefinite integral of e^-x2. Try
u = x
v = x*e(-2amx²/ħ)

 

[BvU]

Hello Lasse,

Dr Claude gave you a good answer. I played with typesetting (a hobby) the question and got to the point where it starts to derail. Pity to waste the effort, so see below.

$$ ... latex code ... $$ creates displayed math (the centered equations) and
$... latex code ...$ creates in-line math
You can check out the $\LaTeX$ code by clicking on an equation with the right mouse button and then pick "show math as $\TeX$ commands"

\left and \right are grouping brackets that grow with the stuff in between.
\; and \; are small spacings

The question is as follows:

A particle of mass m has the wave function
$$\psi(x, t) = A * \exp\left ( -a\left ( { mx^2 \over \hbar} +it \right ) \right )$$
where A and a are positive real constants.

I don't know how to format my stuff on this website, so it may be a bit harder to read. Generally when I write "int" i mean the integral from -infinity to +infinity.
first it asks me to find A, i do this by normalizing the wave function, and i find that
$$A = \left ( 2am \over \pi \hbar \right )^ { 1 / 4}$$
this is correct according to the solutions manual.

I'm then asked to find the potential energy function, but since this doesn't mean anything for the rest of the question i will ignore it.
I am then asked to calculate the expectation values of x, x^2, p and p^2
so starting with the expectation value of x (I will write this as <x>) :

from calculating A I already know that $$\int \exp \left ( -2a\, {mx^2 \over \hbar} \right ) \; dx = \left ( \pi \hbar \over 2am \right )^{ 1 / 2} = A^{-2}$$
$$<x> = \int x \, |\psi(x, t)|^2 \,dx = A^2 \int \, x \exp \left ( -2a { mx^2 \over\hbar } \right ) \; dx $$
now I use integration by parts
$$ \int u \, dv = uv - \int v \, du$$
I use the following u and v: $u = x \Rightarrow du = dx$ (comment: so no need to make that change)
$$dv = \exp\left ( -2 a {mx^2 \over \hbar} \right ) \; dx$$

...

Here things go wrong and you're getting help already.

 

[Lasse Jepsen]

That is not legitimate, because v must be an indefinite integral, and there isn't an analytical form for the indefinite integral of e^-x2. Try
u = x
v = x*e(-2amx²/ħ)

There are a few problems when doing this, integration by parts says:

$$ \int(u dv) = uv - \int(v du) $$

if i use $u = x$ and $v = x e^{-2amx^2 \over \hbar }$ then $dv/dx = e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x$
this way $dv = e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x dx$

making the left side of my integral look like this:

$$ \int(u dv) = \int(x (e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x dx) $$

and this isn't the integral that i am looking for.

if i use $u = x$ and $dv = x e^{-2amx^2 \over \hbar } dx$
then my $v = \int( x e^{-2amx^2 \over \hbar } dx )$
in this case i get a new integral with the exact same problem, because v must be an indefinite integral, and i can't calculate the indefinite integral
$\int( e^{-2amx^2 \over \hbar } dx )$

So I am still completely clueless as to what i should do for this problem.

 

[DrClaude]

So I am still completely clueless as to what i should do for this problem.

I told you above: you have to look up how to perform Gaussian integrals. You can't solve that integral using integration by parts.

 

[mjc123]

Apologies, I meant dv = xe^(-2amx²/ħ)dx. And I believe it does work, provided the integral is from -∞ to ∞.

 

[Ray Vickson]

There are a few problems when doing this, integration by parts says:

$$ \int(u dv) = uv - \int(v du) $$

if i use $u = x$ and $v = x e^{-2amx^2 \over \hbar }$ then $dv/dx = e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x$
this way $dv = e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x dx$

making the left side of my integral look like this:

$$ \int(u dv) = \int(x (e^{-2amx^2 \over \hbar } + -2amxe^{-2amx^2 \over \hbar } x dx) $$

and this isn't the integral that i am looking for.

if i use $u = x$ and $dv = x e^{-2amx^2 \over \hbar } dx$
then my $v = \int( x e^{-2amx^2 \over \hbar } dx )$
in this case i get a new integral with the exact same problem, because v must be an indefinite integral, and i can't calculate the indefinite integral
$\int( e^{-2amx^2 \over \hbar } dx )$

So I am still completely clueless as to what i should do for this problem.

You can immediately conclude that $\langle x \rangle = 0$ without doing any work, just because the probability density function has the form $P(x)= N e^{-\alpha x^2}$ for some constants $\alpha > 0$ and $N > 0.$ The function $P(x)$ is even (meaning that $P(-x) = P(x)$ for all $x$), so the integral $\int x^k P(x) \, dx = 0$ for any odd integer $k$.

For $k = 2$ you can write the integral as
$$\int x^2 P(x) \, dx = \int x \frac{1}{2 \alpha } (2 \alpha x e^{-\alpha x^2} ) \, dx =\frac{1}{2\alpha} \int x \, d\left(-e^{-\alpha x^2} \right) $$

Easier: if you know probability theory you will know that the standard normal distribution function
$$f(x) = \frac{1}{\sqrt{2 \pi} \, \sigma} e^{-x^2/(2 \sigma^2)}$$ has mean 0 and variance $\sigma^2$, meaning that $\int_{R} x^2 f(x) \, dx = \sigma^2.$ Then, all you need to is figure out how the parameter $\sigma^2$ relates to the inputs in your problem.