# Question on Notation

1. Mar 27, 2006

### eep

Hi,
I've seen the Schrodinger equation written in the following form:

$$i\hbar\frac{\partial\Psi}{{\partial}t} = -\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi$$

where

$$\nabla^{2} = \frac{\partial^{2}}{{\partial}x^2} + \frac{\partial^2}{{\partial}y^2} + \frac{\partial^2}{{\partial}z^2}$$

Now, is $\nabla^2\Psi$ a vector or a scalar? In this notation, I would say it's a vector. You have $\nabla^2$ acting on each of the components of $\Psi$. However the book seems to say that $\nabla^2\Psi$ is a scalar. Shouldn't the notation then be $\nabla^2\cdot\Psi$? That is, shouldn't it be a dot product? I'm rather confused...

Last edited: Mar 27, 2006
2. Mar 27, 2006

### nrqed

You seem to thing of $\Psi$ as a 3-D vector with i,j,k component. Thats not the case. It is a scalar function.

Pat

3. Mar 27, 2006

### eep

Can't we represent $\Psi$ as a vector in hilbert space?

4. Mar 27, 2006

### Hurkyl

Staff Emeritus
Yes -- and the vectors in that Hilbert space are the scalar functions. They're not geometric vectors.

(There's a reason you learned about vector spaces other than the n-tuples in your linear algebra course!)

5. Mar 27, 2006

### nrqed

(I *thought* we would be getting to this point but I did not want to muddle the waters too fast). You are perfectly right. Except that this use of the word vector has nothing to do with the usual use of vector as meaning soemthing of the form $A_x {\vec i} + A_y {\vec j} + A_z {\vec k}$. The solutions of Schrodingers equations form a vector space in the more general mathematical sense, but they are each scalar functions.

Pat

6. Mar 27, 2006

### eep

Ah, of course. I was thinking that when I posted but didn't want to jump the gun either. So the Schrodinger equation acts on $\Psi$ which is a function of x,y,z,t and whose range is scalars. Each $\Psi$ lives in Hilbert space and can be represented by a vector, but this has nothing to do with the way the Schrodinger equation acts on $\Psi$, right?

7. Mar 28, 2006

### reilly

DEL**2 is a scalar (under rotations). This follows from a standard vector calculus convention, that DEL or GRADIENT is a vector with the following components (d/dx,d/dy/d/dz). Discussed in thousands of text books,
Regards,
Reilly Atkinson

8. Mar 28, 2006

### gulsen

I have another question about notation in QM.
If $$< \Psi | \hat{H} \Psi>$$ is the same thing as $$< \Psi | \hat{H} | \Psi>$$, why do they invent the extra | between?

9. Mar 28, 2006

### ZapperZ

Staff Emeritus
Because that is only valid for a hermitian operator. If you replace that with a non-hermition operator in a purely mathematical exercise, then these two are no longer identical.

Zz.

10. Mar 28, 2006

### Perturbation

For one, it's in coordinate representation, so the wave-function is a scalar given by $\langle x|\psi\rangle$. Here the bra and ket vectors themselves aren't vectors in the usual sense (an n-tuple), they're members of a Hilbert space, an infinite dimensional linear vector space over $\mathbb{C}$, where the members of the space are functions (which obviously satisfy the vector space axioms).

Last edited: Mar 28, 2006
11. Mar 28, 2006

### Physics Monkey

It is really a notational choice and not much more (why keep the extra bar). You can define the ket $$| H \psi \rangle$$ as $$\hat{H} | \psi \rangle$$. What you cannot do is write $$\langle \psi | \hat{H} | \psi \rangle = \langle H \psi | \psi \rangle$$ unless $$\hat{H}$$ is self adjoint. In fact, $$\langle H^\dag \psi | \psi \rangle = \langle \psi | H \psi \rangle = \langle \psi | \hat{H} | \psi \rangle$$ where as $$\langle H \psi | \psi \rangle = \langle \psi | H^\dag \psi \rangle = \langle \psi | \hat{H}^\dag | \psi \rangle$$

Last edited: Mar 28, 2006