# Matrix or differential operator?

1. May 20, 2010

### 1Kris

Hi,
I've been reading a couple of different books on quantum mechanics and have come a mathematical difficulty. I understand that the Hamiltonian is an operator but in some books, it represented as a matrix and in others as a differential operator? How can they both be equivalent approaches?

I understand that they have some of the same properties, eg. are non-commutative. But how can both give the same results? I think (although I may be wrong) that I am asking for the relations between a function and a vector.

Thanks

2. May 20, 2010

### Fredrik

Staff Emeritus
See this post (or any book on linear algebra) for the relationship between matrices and linear operators between finite-dimensional vector spaces.

3. May 20, 2010

### 1Kris

Thanks for the speedy and clear response. Does that suggest that any linear operator A can be written as a matrix and if so is there a general method?

4. May 20, 2010

### Fredrik

Staff Emeritus
Yes it does, when the vector spaces are finite-dimensional. The stuff I wrote about there is the method. I decided to edit my post, but you replied while I was typing so I'm posting the edited version here instead:

See the quote below (or any book on linear algebra) for the relationship between matrices and linear operators between finite-dimensional vector spaces.
Also, consider the special case U=V, with an orthonormal basis $\{u_i\}$. (Set $v_i=u_i$).

$$\langle u_i,Au_j\rangle=\langle u_i,(Au_j)_k u_k\rangle=(Au_j)_k\langle u_i,u_k\rangle=(Au_j)_k\delta_{ik}=(Au_j)_i=A_{ij}$$

Last edited: May 20, 2010
5. May 20, 2010

### jostpuur

Using discrete approximations may help seeing what's going on.

Suppose you are interested in some functions $f:[0,1]\to\mathbb{C}$.

You can approximate the interval $[0,1]$ as a discrete set

$$\big\{0,\frac{1}{N},\frac{2}{N},\ldots,\frac{N-1}{N}\big\}$$

with some large $N$. Then the functions $f:[0,1]\to\mathbb{C}$ can approximated as functions

$$f:\big\{0,\frac{1}{N},\frac{2}{N},\ldots,\frac{N-1}{N}\big\}\to\mathbb{C}.$$

But now these new functions are merely vectors in N-dimensional vector space. Like

$$(v_1,v_2,\ldots, v_N) \approx \Big(f(0), f\big(\frac{1}{N}\big), f\big(\frac{2}{N}\big),\ldots, f\big(\frac{N-1}{N}\big)\Big)$$

Any operator that was form $A:D(A)\to \mathbb{C}^{[0,1]}$ with some domain $D(A)\subset \mathbb{C}^{[0,1]}$, can now be approximated as a $N\times N$-matrix.

This approach gives some insight into what it means that function spaces are infinite dimensional vector spaces.

6. May 20, 2010

### 1Kris

So roughly speaking are we saying that each value of a function can be thought of as one of infinitely many components of a vector?

7. May 20, 2010

### LostConjugate

Yes the argument becomes the index and the value of the function at that argument becomes the component of that index.

8. May 21, 2010

### Edgardo

Last edited by a moderator: May 4, 2017