2013-02-28 14:50:41 8 Comments

Would anyone be able to explain the difference, technically, between wave function notation for quantum systems e.g. $\psi=\psi(x)$ and Dirac bra-ket vector notation?

How do you get from one to the other formally?

When you express a state in bra ket notation as a vector $(a,b)$ for instance do we have to be referring to a basis of eigenvectors for an observable?

Furthermore, consider if we had a unitary function $U$ how would we express $\langle U \psi\mid$ in terms of $\langle \psi\mid$? As the bra of psi is an element of the duel Hilbert space, which is a function that takes the ket of psi to the inner product, how would we remove the unitary operator?

Note, this is not a homework question, i'm just trying to improve my formal understanding of the notation I've been using, as this has been skirted over quite substantially.

EDIT: Got the last question I asked $$\langle U\psi \mid \phi \rangle=\langle \psi \mid \bar{U^T}\phi \rangle$$ $$\forall \phi \in H \Rightarrow \langle U\psi\mid = \langle \psi\mid\bar{U^T}.$$

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## 1 comments

## @joshphysics 2013-02-28 15:22:18

Here's how you get from one to the other. Let me take the case of a particle moving in one dimension.

In this case, we assume that there exist vectors $|x\rangle$ which form a "dirac-normalized" basis (the position basis) for the Hilbert space in the sense that their inner products satisfy $$ \langle x|x'\rangle =\delta(x-x') $$ Note, as an aside, that these vectors are not normalizable in the standard sense, and therefore they do not strictly speaking belong to the Hilbert space.

Next, for each $|\psi\rangle$ in the Hilbert space, we define the position basis wavefunction $\psi$ corresponding to the state $|\psi\rangle$ as $$ \psi(x) = \langle x|\psi\rangle $$ So really, the value $\psi(x)$ of the position basis wavefunction $\psi$ at a point $x$ can simply be thought of as the basis component of $|\psi\rangle$ in the direction of $|x\rangle$ just as in the finite-dimensional case where one can find the component of a vector $|\psi\rangle$ along a basis vector $|e_i\rangle$ simply by taking the inner product $\langle e_i|\psi\rangle$.

## @Freeman 2013-02-28 15:36:17

This position basis you refer to, are they not normalizable in the standard sense because they are related to this: $X\psi=x\psi=\alpha \psi \Rightarrow (x-\alpha)\psi=0$, is the basis $\mid x \rangle$ this set of $\psi$? I'm struggling slightly to see how $\psi(x) = \langle x|\psi\rangle$, sorry i'm being a bit slow, but i'm used to working with wave functions and when I see that everything is telling me i'm taking an integral over the whole domain. Thanks so much for the help though, I really appreciate it!

## @joshphysics 2013-02-28 16:03:15

Yeah so, in the position basis, the position operator acts by multiplication by the argument of the function, $X\psi(x) = x\psi(x)$, so if $\psi_{x'}(x)$ denotes an eigenvector of $X$ with eigenvalue $x'$, namely $X\psi_{x'}(x) = x'\psi(x)$, we get $x\psi_{x'}(x) = x'\psi_{x'}(x)$ which can only hold if $\psi(x)=0$ for all $x\neq x'$. In other words $\psi_{x'}$ must be zero at all but one point; in particular it can't be normalized to $1$. Therefore, it is not an element of the Hilbert space, it is, in fact, the Dirac delta distribution which is another mathematical beast altogether.

## @Freeman 2013-02-28 16:07:01

Excellent, thank you, that's what I thought. You have been very helpful :)

## @joshphysics 2013-02-28 16:12:40

Great! I'm glad it's more clear now. I'm guessing you're using Griffiths, and I know I always felt he did a poor job of describing Dirac notation. Cheers!

## @Freeman 2013-02-28 16:15:17

I'm using KC Hannabuss actually, but I've just started quantum information theory so wanted to solidify this in my head :) Thanks so much, I really appreciate your time.

## @joshphysics 2013-02-28 17:01:37

Oh interesting you're a math student. I was pure math as well and used that book in an applied maths class. You might find this amazon.com/gp/product/0821846302/… helpful...I did.

## @Freeman 2013-03-01 14:10:38

Indeed I am :) Oh really, it's a great book. Thanks for the recommendation, that's very handy!