# Quantum Mechanics

• Dec 9th 2007, 08:26 AM
albi
Quantum Mechanics
I have the following problems:

a) Under which conditions we can simultaneously measure (with arbitrary precision) two observables which operators does not commute?

b) Does energy and coordinates of particle's momentum are in general simultaneously measurable. Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property.

c) Are there any conditions under which position and square of momentum are simultaneously measurable. Find an example of such conditions.
• Dec 9th 2007, 08:42 AM
albi
My ideas
a) The uncertainity principle goes like that:

$\displaystyle \sigma^2(A) \sigma^2(B) \geq \frac{1}{4} \left| \langle \psi^* | [A, B] | \psi \rangle \right|^2$

So I think the right answer is: "The observables are simultaneously measurable if mean $\displaystyle \langle \psi^* | [A, B] | \psi \rangle$ is zero."
• Dec 9th 2007, 09:17 AM
albi
my ideas II
b)

Let $\displaystyle D_i$ denote differentiation over i-th coordinate. We have one particle with mass m. The energy operator should be like that:
$\displaystyle E = -\frac{\hbar^2}{2m} \nabla^2 = -\frac{\hbar^2}{2m} \sum_i D^2_i$

Momentum operator (coordinate k):

$\displaystyle p_k = -i\hbar D_k$

I will focus on the commutator: $\displaystyle [p_k, E]$

$\displaystyle [p_k, E] = \frac{i\hbar^3}{2m}\left[D_k, \sum_i D^2_i \right] = \frac{i\hbar^3}{2m}\sum_i [D_k, D^2_i]$

I shall assume $\displaystyle i \neq k$:

$\displaystyle [D_k, D^2_i] = D_kD^2_i - D^2_iD_k$

Let's assume the wavefunction belongs to $\displaystyle C^\infty$ class (mathematiclly very restrictive assumption but I consider it as "physical")

As I remember there was some theorem which states that we can change sequence of differentiation (I guess the assumption was that the second derivative is continuous). It was important theorem but I don't remember it's name :( .

So under this assumptions for wavefunction we should get:

$\displaystyle D_kD^2_i \psi = D_kD_iD_i \psi = D_iD_kD_i \psi = D_iD_iD_k \psi = D^2_iD_k \psi$

So the commutator $\displaystyle [D_k, D^2_i]$ is zero

I guess it is obvious that: $\displaystyle [D_k, D^2_k] = 0$

So $\displaystyle [p_k, E] = 0$

Which means that operators $\displaystyle p_k$, $\displaystyle E$ commute.

Thus we can simulataneously measure momentum and energy with infinite precision.

It seems to me a little strange because the question "Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property." suggests that they should not commute (otherwise giving the example is trivial - any system has such proprety). Could you check if my reasoning is correct?
• Dec 9th 2007, 09:30 AM
topsquark
Quote:

Originally Posted by albi
a) The uncertainity principle goes like that:

$\displaystyle \sigma^2(A) \sigma^2(B) \geq \frac{1}{4} \left| \langle \psi^* | [A, B] | \psi \rangle \right|^2$

So I think the right answer is: "The observables are simultaneously measurable if mean $\displaystyle \langle \psi^* | [A, B] | \psi \rangle$ is zero."

Yes.

-Dan
• Dec 9th 2007, 09:33 AM
topsquark
Quote:

Originally Posted by albi
b)

Let $\displaystyle D_i$ denote differentiation over i-th coordinate. We have one particle with mass m. The energy operator should be like that:
$\displaystyle E = -\frac{\hbar^2}{2m} \nabla^2 = -\frac{\hbar^2}{2m} \sum_i D^2_i$

Momentum operator (coordinate k):

$\displaystyle p_k = -i\hbar D_k$

I will focus on the commutator: $\displaystyle [p_k, E]$

$\displaystyle [p_k, E] = \frac{i\hbar^3}{2m}\left[D_k, \sum_i D^2_i \right] = \frac{i\hbar^3}{2m}\sum_i [D_k, D^2_i]$

I shall assume $\displaystyle i \neq k$:

$\displaystyle [D_k, D^2_i] = D_kD^2_i - D^2_iD_k$

Let's assume the wavefunction belongs to $\displaystyle C^\infty$ class (mathematiclly very restrictive assumption but I consider it as "physical")

As I remember there was some theorem which states that we can change sequence of differentiation (I guess the assumption was that the second derivative is continuous). It was important theorem but I don't remember it's name :( .

So under this assumptions for wavefunction we should get:

$\displaystyle D_kD^2_i \psi = D_kD_iD_i \psi = D_iD_kD_i \psi = D_iD_iD_k \psi = D^2_iD_k \psi$

So the commutator $\displaystyle [D_k, D^2_i]$ is zero

I guess it is obvious that: $\displaystyle [D_k, D^2_k] = 0$

So $\displaystyle [p_k, E] = 0$

Which means that operators $\displaystyle p_k$, $\displaystyle E$ commute.

Thus we can simulataneously measure momentum and energy with infinite precision.

It seems to me a little strange because the question "Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property." suggests that they should not commute (otherwise giving the example is trivial - any system has such proprety). Could you check if my reasoning is correct?

You are working at it too hard. $\displaystyle [ D_i, D_k ] = 0$ for all i and k, practically by definition.

Give me a bit to think of a system where you might run into a problem. (And remember, your energy function is not quite general: you could have a potential function in there...)

-Dan
• Dec 9th 2007, 10:05 AM
albi
My ideas III
c)

Again I will consider commutator $\displaystyle [x, p^2]$ in one dimensional space (for simplicity).

We have:

$\displaystyle p^2 = -\hbar^2 \frac{d^2}{dx^2}$

So:

$\displaystyle [x, p^2] = -\hbar^2 \left[x, \frac{d^2}{dx^2}\right]$

$\displaystyle \left[x, \frac{d^2}{dx^2}\right] = x\frac{d^2}{dx^2} - \frac{d^2}{dx^2} x$

Let $\displaystyle \psi$ be a wavefunction. We have:

$\displaystyle \frac{d^2}{dx^2} x \psi = \frac{d}{dx} \frac{d}{dx}x\psi = \frac{d}{dx} \left ( \psi + x \frac{d\psi}{dx} \right) = \frac{d\psi}{dx} + \frac{d\psi}{dx} + x \frac{d^2\psi}{dx^2}$

Hence:
$\displaystyle \frac{d^2}{dx^2} x = 2 \frac{d}{dx} + x \frac{d^2}{dx^2}$

Thus:
$\displaystyle \left[x, \frac{d^2}{dx^2}\right] = -2 \frac{d}{dx}$

Finally:

$\displaystyle [x, p^2] = 2\hbar^2 \frac{d}{dx}$

We should have: $\displaystyle \langle \psi^* | [x, p^2] | \psi \rangle = 0$

It is equivalent to:

$\displaystyle 2\hbar^2 \int_{-\infty}^{\infty} \psi^*(x) \frac{d\psi(x)}{dx} dx = 0$

I cannot think of any $\displaystyle \psi$ having such property.
• Dec 9th 2007, 10:06 AM
albi
Quote:

Originally Posted by topsquark
Give me a bit to think of a system where you might run into a problem. (And remember, your energy function is not quite general: you could have a potential function in there...)

-Dan

I know, but I wanted to consider the simplest case for the beginning.
• Dec 9th 2007, 10:08 AM
albi
Quote:

Originally Posted by albi
I know, but I wanted to consider the simplest case for the beginning.

Or maybe I should consider more general case. Maybe $\displaystyle p_k$ and $\displaystyle E$ would not commute if I add the potential?
• Dec 9th 2007, 10:26 AM
albi
So I got

$\displaystyle [p_k, E] = \frac{-i \hbar^3}{2m} V_k(\mathbf{x})$

Where $\displaystyle V_k = D_k V$

The most trivial case is when: $\displaystyle V = 0$. Maybe the good example would be "infinite potential well" ?
• Dec 9th 2007, 10:35 AM
albi
Problem c)

I guess i can do that:

$\displaystyle \int_{-\infty}^{\infty} \psi^*(x) \frac{d\psi(x)}{dx} dx = \int z^* dz$

So the problem is to find the curve $\displaystyle \gamma$ for which: $\displaystyle \int_{\gamma} z^* dz = 0$

But still I don't have idea :(
• Dec 9th 2007, 05:59 PM
topsquark
Quote:

Originally Posted by albi
Finally:

$\displaystyle [x, p^2] = 2\hbar^2 \frac{d}{dx}$

We should have: $\displaystyle \langle \psi^* | [x, p^2] | \psi \rangle = 0$

It is equivalent to:

$\displaystyle 2\hbar^2 \int_{-\infty}^{\infty} \psi^*(x) \frac{d\psi(x)}{dx} dx = 0$

I cannot think of any $\displaystyle \psi$ having such property.

Then let's keep it simple: Consider $\displaystyle \psi(x) = sin(x)$.

-Dan
• Dec 9th 2007, 06:04 PM
topsquark
Quote:

Originally Posted by albi
So I got

$\displaystyle [p_k, E] = \frac{-i \hbar^3}{2m} V_k(\mathbf{x})$

Where $\displaystyle V_k = D_k V$

The most trivial case is when: $\displaystyle V = 0$. Maybe the good example would be "infinite potential well" ?

Actually
$\displaystyle [ p_k, E ] \implies [p_k, p^2 + V] = [p_k, V]$

So yes, a constant potential will work for this. So will a step potential, and possibly a delta function potential (don't quote me on this one, I haven't checked it.)

-Dan
• Dec 10th 2007, 12:43 AM
albi
Quote:

Originally Posted by topsquark
Then let's keep it simple: Consider $\displaystyle \psi(x) = sin(x)$.

-Dan

That was my first thought, but I rejected it because there is problem with integral (it is not convergent)...

But if we have sin(x) on some finite interval (integer number of periods) than it should work :). Thanks for help.
• Dec 10th 2007, 06:13 AM
topsquark
Quote:

Originally Posted by albi
That was my first thought, but I rejected it because there is problem with integral (it is not convergent)...

But if we have sin(x) on some finite interval (integer number of periods) than it should work :). Thanks for help.

Yeah, sin(x) is a lousy wavefunction, isn't it?.I was keeping things too simple. Let's be more general and say that any function $\displaystyle \psi(x)$ that is continuous and bounded with $\displaystyle \lim_{x \to \pm \infty}\psi(x) = 0$ that is an even function will do the trick.

-Dan