(Quantum) linear operators

Just starting third level Uni. stuff & am faced with linear operators from Quantum Mechanics & need a little help.

OK, an operator, Ô, is said to be linear if it satisfies the equation

Ô(α f1 + β f2) = α(Ô f1) + β(Ô f2)

Fine

but I have an equation I can't wrap my head around, maybe just rusty, a hint would be nice, though.

Ô1 = d/dx;

Ô2 =3 d/dx +3x^2;

Find the new functions obtained by acting with each of these operators on

(a) g(x, t) =3 d/dx +3x^2

(b) h(x, t)=α sin(kx − ωt).

Now

Ô1 g(x,t) = 6xt^3

But not sure about how to get

Ô2 g(x,t) =

how to get this middle bit, please . . . . .

Answer is 18xt^3 + 9x^4 t^3

Re: (Quantum) linear operators

it appears you have not given a proper listing for g(x,t), as you just have the definiton of $\displaystyle \hat{O}_2$ repeated there.

it is also unclear how "x" is supposed to operate on a function g(x,t).

Re: (Quantum) linear operators

Quote:

Originally Posted by

**froodles01** Just starting third level Uni. stuff & am faced with linear operators from Quantum Mechanics & need a little help.

OK, an operator, Ô, is said to be linear if it satisfies the equation

Ô(α f1 + β f2) = α(Ô f1) + β(Ô f2)

Fine

but I have an equation I can't wrap my head around, maybe just rusty, a hint would be nice, though.

Ô1 = d/dx;

Okay, that is "differentiate the function".

This says "differentiate the function, then add x^2 and multiply by 3.

Quote:

Find the new functions obtained by acting with each of these operators on

(a) g(x, t) =3 d/dx +3x^2

This is **not** a function. With that "d/dx" in it, it is another operator. Further, there is no "t" in it. In fact, this is the same as your definition of Ô2. I suspect you have copied the wrong thing.

Quote:

(b) h(x, t)=α sin(kx − ωt).

Now

Ô1 g(x,t) = 6xt^3[/quote\

This would be correct if g(x,t) were just 3x^2t^3 or that plus a constant.

But not sure about how to get

Ô2 g(x,t) =

Take your previous answer, which was apparently 6xt^3, add x^2, and multiply by 3: 3(6xt^3+ x^2)= 18xt^3+ 3x^2.

Quote:

how to get this middle bit, please . . . . .

Answer is 18xt^3 + 9x^4 t^3

I think you need to go back, read the problem again and see if you have not copied the first function incorrectly.

1 Attachment(s)

Re: (Quantum) linear operators

Hmmm. I have scanned the relevant bit in & attached it here.

Aha! Yes, the devil's in the detail, of course & that's quite a big detail.

Thank you for the heads up!

Attachment 26767Attachment 26767