# A little Bra-Ket notation theorem that I don't get

1. Feb 27, 2013

### Excimer

I'm continuing through Dirac's book, The Principles of Quantum Mechanics. You can view this as a google book in the link below.

On page 28-29 he proves this Theorem:

If ξ is a real linear operator and

ξm|P> = 0 (1)

for a particular ket |P>, m being a positive integer, then

ξ|P> = 0

...

And the proof is fine by me except for the first part where he says:

put m = 2 then (1) gives

<P|ξ2|P> = 0 (2)

and goes on to prove it by induction. But I don't see how (2) is 'given' from (1).

Can anyone explain?

If he is relying on assuming (1) is true for some ket |P> and positive integer m, then I just pick m = 1 and it's not a theorem. And besides, in this case he never demonstrates a definite example of the statement being true, which you need to do in induction.

Last edited by a moderator: May 6, 2017
2. Feb 27, 2013

### Excimer

By the way I do understand that because ξ is a real linear operator that the conjugate to ξ|P> is <P|ξ so that you can write <P|ξξ|P>, but I don't see that it follows this is equal to zero.

3. Feb 27, 2013

### micromass

Staff Emeritus

Are there any conditions on $\xi$? If you only assume it to be a real linear operator, then I think I can find counterexamples to the statement.

4. Feb 27, 2013

### Excimer

5. Feb 27, 2013

### CompuChip

The assumption is that for some positive integer m, equation (8) holds.
Clearly, if m = 1 the statement is trivial, so he considers m = 2 as the base case.

So first he proves that if ξ²|P> = 0, then ξ|P> = 0. The reasoning is as follows: assume ξ²|P> = 0. <P| is a linear operator, so <P|0 = 0. Hence, <P|ξ²|P> = 0. However, instead of looking at that expression as <P| applied to ξ²|P>, you can also view it as <P|ξ applied to ξ|P>. Since ξ is real, <P|ξ = <P|ξ*, so what you have is really || ξ|P> ||² (the length of ξ|P>) which can only be zero if ξ|P> is zero.

Then he shows that if it is not given that ξ²|P> = 0 but ξm|P> = 0 for some higher m, it can be reduced to the m = 2 case from which we have just shown that the m = 1 case follows.

6. Feb 27, 2013

### micromass

Staff Emeritus
That is not really true, is it??

Anyway, a counterexample is given by

$$A=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)$$

Then $A^2 = 0$, but $A\neq 0$. So take, for example, |P> = (0,1) and the theorem is false.

7. Feb 27, 2013

### CompuChip

Good point.
It's been a while, wasn't it ξ|P> = <P|ξ?
So you need A = A := (A*)tr, in other words, for ξ to be Hermitian.

8. Feb 27, 2013

### Excimer

Compuchip,

So you mean to say that the theorem should specify m > 1 since otherwise you allow the trivial solution? It might be a silly detail but I find things more confusing when they are not stated rigorously.

In any case I think I get it now thanks to you.

9. Feb 27, 2013

### CompuChip

It doesn't have to: it is also true for m = 1, but trivially so, as you say. However, if you want to prove it for all m then this case doesn't help you, so you need to consider m = 2 as the base case.

10. Feb 27, 2013

### micromass

Staff Emeritus
Yes, you need $\xi$ to be Hermitian. Or since $\xi$ is already real, you need it to be symmetric.
Or a little more general, if $\xi$ is normal (= commutes with its adjoint), then it holds. But it fails for more general operators.

11. Feb 27, 2013

### micromass

Staff Emeritus
Oh, apparently Dirac uses the term "real linear operator" to mean "self-adjoint operator". In that case, the theorem holds true, of course. But it's a weird terminology...

12. Feb 27, 2013

### Excimer

If ξ is real then it equals it's own adjoint, and on page 28 Dirac remarks that if ξ is real than so is ξξ, so if you can take that at face value then ξ must commute with it's adjoint.

13. Feb 27, 2013

### Excimer

Ah, you got in before me.

Ok thanks both of you for helping me out.

Cheers

14. Feb 27, 2013

### dextercioby

Hi micromass,

Dirac wrote his QM book in 1930, before the terminology imposed in functional analysis through M.H.Stone's 1932 book appeared. He did review his QM book and 3 more editions appeared, but the mathematical terminology was not adjusted.