# ^, the operators in quantum mecħanics

1. Oct 12, 2008

### shanu_bhaiya

The doubt:
It's not a problem, but a doubt. We know that in general quantum physics at undergraduate level, we write pΨ = (ħ/i) dΨ/dx. My doubt is that if we derived this equation from Schrodinger's equation only, so we must operate p on a wave-function only, which satisfies Schrodinger's equation.

But I went further, and I saw a problem in the book - "Concepts of Modern Physics - A. Beiser", Chapter 5, Problem 9. It was asked to find the value of <xp>-<px>. Now to solve the problem, p is operated on xΨ (for <px>) and Ψ (for <xp>) both simultaneously. But I can prove that if Ψ satisfies Schrodinger's equation, xΨ cannot. So how can we operate p on xΨ, when it doesn't satisfy the Schrodinger's equation.
Or may be p can operate on anything, then we need a proof, which I may have not yet studied, is it true that there exists such a proof?

2. Oct 12, 2008

### borgwal

Good question: $\hat{p}$ is meant to operate on any wavefunction with suitable properties (such as being square integrable and differentiable), but being a solution to the S.E. is not one of those properties! Otherwise it would even be hard to define $\hat{p}^2$ acting on the wavefunction, since $\hat{p}\psi$ is also, in general, no longer a solution to the S.E.

3. Oct 12, 2008

### shanu_bhaiya

But we derived the meaning of operator p from SE:

Ψ = Aexp((i/ħ)xp)
Differentiate w.r.t. x
dΨ/dx = ((i/ħ )p)Aexp((i/ħ)xp)
put value of Ψ from first equation
dΨ/dx = ((i/ħ )p)Ψ
pΨ = (ħ/i) (dΨ/dx)

Now, 1st equation is nothing but SE, as it's differential form is SE after eliminating A.

Actually, if Ψ is a solution, dΨ/dx is also a solution of SE as you can check. Further, $\hat{p}^2$ can still be derived as follows:

From the last equation above
pΨ = (ħ/i) (dΨ/dx)
Differentiaite it again w.r.t. x
p(dΨ/dx) = (ħ/i) (d2Ψ/dx2)
Replace the value of dΨ/dx from the equation of p operator
p(pΨ/(ħ/i)) = (ħ/i) (d2Ψ/dx2)
p2Ψ = (ħ/i)2 (d2Ψ/dx2)

And hence we can get the value of $\hat{p}^2$. Waiting for your reply...

4. Oct 12, 2008

### borgwal

No, $\hat{p}\psi$ is *not* a solution for the S.E. with a spatially dependent potential.

The meaning of the operator d/dx can be made plausible by looking at a particular wavefunction exp(ikx), but that doesn't mean it is derived from there. In fact, what one requires is the commutation relation $[\hat{x},\hat{p}]=i\hbar$, and from that relation it follows what $\hat{p}$ is.

In any case, you seem to think that the S.E. does not contain a potential....but it does!

5. Oct 13, 2008

### Gokul43201

Staff Emeritus
No, that's not true at all. In fact, the momentum operator used in the SE can be different in different circumstances (see, for example, charged particle in a magnetic field).

The QM momentum operator is "derived" by extension of the classical case, using the fact that momentum is the generator of infinitesimal translations.

See:
1. Goldstein, Classical Dynamics
2. Sakurai, QM

6. Oct 13, 2008

### shanu_bhaiya

Thanks to both of you for clearing the doubt.

7. Nov 7, 2008

### String_man

Let’s make it simple. If I am right, your question is pointing toward something which says,
“will a wave function, that is satisfying the SE (say Ψ) will remain a perfect wave
function when multiplied by something say ‘x’, and thus making it ‘xΨ’ ”? This question is
however a complete analog of something which says, “if we have a continuous,
differentiable and single valued function f(x), will it remain pursuing these properties
even if is multiplied by some other function say g(x)? ”
K(x) = f(x) * g(x)
In case of our analogy, it depends on the continuity and differentiability of g(x), if it is,
then K(x) would be, or otherwise. Fine, but in case of our wave function the thing goes
quite differently, there are so many propert ies like normalization, single valued etc etc etc,
that a function has to satisfy in order to be a perfect wave function . ‘p’, however should
only to be applied to a perfect wave func tion (as it also have to satisfy SE, which is only
possible for a perfect wave function).
Now, taking equation <px>–<xp> which is nothing but the uncertainty principle can be
written as
<ħ (∂x/∂x)/i> – <xħ(∂/∂x)/i>
=> <ħ/i> – <0>
=> <ħ/i> = ħ/i
This is a completely safe, in the sen se that we don’t need to operate the whole equation
on to a wave function kind of stuff (Ψ). Which is simply because the expectation values
of <px> or of <xp> yield no operator, and it ’s operator which operates.
<px>Ψ or <xp>Ψ is wrong, and so as ‘p*x Ψ’.
I think, perhaps you’ll get it now.

8. Nov 7, 2008

### shanu_bhaiya

It is sure that xΨ won't obey SE, you can confirm it by putting in the SE. You'll get some extra terms apart from SE.

Well, whatever... the main point was that the value for p-operator wasn't derived from SE, as Gokul43201 explained that p-operator has been derived by extension of the classical case using the fact that it is the generator of the infiniteimal translations. Although I need more insight in this.

9. Nov 7, 2008

### Ben Niehoff

For any general linear operator A, to say that it is the "generator" of some group action G, means this:

$$e^A = G$$

where the operator exponential is defined by the infinite series

$$e^A = 1 + A + \frac{1}{2} A^2 + \frac{1}{3!} A^3 + ...$$

if you put the derivative operator in place of A in the above series, you can check that it is, indeed, the generator of translations.

Note: The factor $-i \hbar$ is used to make the momentum operator Hermitian (such that its inverse is its conjugate transpose). In this case, the infinitesimal translation operator is

$$T \psi(x) = \psi(x - \frac{i}{\hbar} \epsilon)$$

rather than

$$T \psi(x) = \psi(x - \epsilon)$$