# Math step in Dirac's book

1. Jan 17, 2014

### Catflap

In Principles of QM by Dirac, 4th Ed page102 he gives the following:-

Limδx→0(De-1)/δx = Limδx→0(D-1+iγ)/δx
given that the phase factor tends to 1 in the limit

I'm stuck here. My grasp of maths is pretty weak and I can't see how the pure imaginary number arises. Any help would be appreciated.

2. Jan 17, 2014

### fobos3

From Euler's formula:
$$e^{i\gamma} = \cos\gamma + i\sin\gamma$$
If the exponential approaches 1, then gamma approaches 0. Now:
$$\cos 0 = 1$$
Also, when gamma is small:
$$\sin\gamma \approx \gamma$$
Which you can get from the Taylor series. Finally:
$$e^{i\gamma} \approx 1 + i\gamma$$
I suppose you can argue that:
$$D\gamma \approx \gamma$$
for small gamma.

3. Jan 17, 2014

### Catflap

That's the problem. D is an operator, so I have D+iDγ - not the same as D + iaγ where a is pure number.

4. Jan 17, 2014

### fobos3

Is D like a specific operator(differential operator?) or is it any operator?

5. Jan 17, 2014

### Catflap

It's an operator related to displacement such that a displaced ket relates to the original ket by some operator D. i.e. |Pd> = D|P>.

If δx is an infinitesimal displacement from the initial position, then from physical continuity a displaced ket should tend to the original ket so that the limit limδx→0 (|Pd> -|P>)/δx should exist.
That can be written limδx→0(D-1)/δx|P> = dx (the displacement operator)

There is an arbitrary phase factor to multiply into D which must also tend to unity as δx tends to 0. That gives the expression as I wrote it in the question. It's the next step in his analysis that I don't understand.

6. Jan 17, 2014

### Staff: Mentor

D is an operator but D can be multiplied by an arbitrary phase factor e^iy (i enters into it because that is the definition of a phase factor, and it is one of the principles of QM you can do this as Dirac explains in earlier pages) with y real and y going to 0 as ∂x goes to zero because D goes to 1.

From the elementary properties of the exponential if y small e^iy = 1 + iy.

So you have D replaced by De^iy because of that arbitrary phase factor, and since y is small if ∂x is small, its replaced by D*(1 +iy) = D + Diy when ∂x small.

Thus (De^iy - 1)/∂x = (D-1)/∂x + D*(iy/∂x).

As ∂x goes to zero (D-1)/∂x goes to dx, D goes to 1 and y/∂x is assumed to go to ax.

Thus you get, in the limit dx + iax.

Just as a comment about learning QM from Diracs classic, I dont advise it.

I did it and it created a number of issues it took me a while to work through, and some only came to light once I started posting here.

A MUCH better book is Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

Once you understand Ballentine then return to Dirac and see all the 'issues' it has. Still a classic of course and should be in the library of anyone interested at a serious level in QM - just not the best place to learn it from.

Also I get the impression your math background may not be good. If that's the case I suggest a more gentle easing into it starting with Lenny Susskinds Books:
https://www.amazon.com/Theoretical-...NBE_1_1?s=books&ie=UTF8&qid=1390002146&sr=1-1
https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465036678

Then Hughes - Structure and Interpretation of Quantum Mechanics:
https://www.amazon.com/The-Structure-Interpretation-Quantum-Mechanics/dp/0674843924

That book is unique at the beginning level in that it covers the very important Gleasons Theorem which is not usually encountered until advanced studies.

That is probably good enough background for Ballentine.

Thanks
Bill

Last edited: Jan 17, 2014
7. Jan 18, 2014

### Catflap

I see, The real part of the expression involves D-1 the imaginary part involves D multiplied by some factor

With the result of the operation D being close to unitary, the real part becomes some small fraction dominated by D, while the imaginary part becomes something dominated by the other factor.

Fair enough.

Thanks for the reading list - I have read Feynman, Griffiths and others, I'm just reading Dirac out of historical interest - I wanted to see how he and others thought about the subject at the time it was developing.

PS Limδx→0(Deiγ-1)/δx = Limδx→0(D-1+iγ)/δx as an equality is wrong then really - it should be Limδx→0(D-1+iDγ)/δx and then you can refine the limit from that position.
I suppose the argument is that the limit is the same and therefore the equality applies.

Last edited: Jan 18, 2014
8. Jan 19, 2014

### forcefield

Where do you get that D goes to 1 ? Isn't it the phase factor that must be made to tend to unity as δx goes to zero, and that's why γ must be made to go zero as δx goes to zero ?

9. Jan 19, 2014

### strangerep

It's part of physical requirements of D, explained by Dirac on p102 and earlier.

This sort of thing is also yet more evidence that learning of QM from Dirac's book is difficult. He's not the best teacher, even though he was a brilliant physicist.

10. Jan 20, 2014

### forcefield

I am reading the book. It says on p102: "The arbitrary numerical factor e with γ real which we may multiply into D must be made to tend to unity as δx → 0". It does not say that D goes to 1.

11. Jan 20, 2014

### Staff: Mentor

D is the displacement operator ie how a state transforms if you displace the measurement apparatus by a distance x. Don't displace it - lo and behold - nothing happens so D = 1.

The reason you can multiply by an arbitrary phase factor is states are physically the same if you do that. Although Dirac doesn't do it, the correct interpretation of a state is as a positive operator of unit trace. He basically only deals with so called pure states which are projection operators |x><x| - and are invariant to multiplying x by a phase factor. Dirac however justifies it on the basis of the principle of superposition - but in modern times that's not the correct way of going about it. Just one reason, classic though that text is, why its not the best place to learn QM - there are things you must unlearn when you see a correct treatment such as Ballentine.

Thanks
Bill

12. Jan 20, 2014

### Staff: Mentor

D must go to 1 by its physical interpretation, regardless of if Dirac explicitly states it or not. Its the reason y must go to zero as ∂x goes to zero - not the other way around.

Thanks
Bill

Last edited: Jan 20, 2014
13. Jan 20, 2014

### forcefield

Where do you get that D is the displacement operator ? It says that dx is the displacement operator.

14. Jan 20, 2014

### Staff: Mentor

Page 101 where it is defined. dx is the derivative of the displacement operator.

Although I understand your confusion - Dirac calls dx the displacement operator - but that's not really the best terminology - its really the derivative of the the operator D that physically displaces a state. At the top of page 102 he calls it the operator that displaces a state - so it's really inconsistent with his later definition of dx, which is on the same page.

Sorry for the confusion - its been years since I read that book and had forgotten the exact terminology he used.

Thanks
Bill

Last edited: Jan 20, 2014
15. Jan 20, 2014

### forcefield

dx = $^{lim}_{δx → 0}$(D-1)/δx

The book says that "this limit is a linear operator which we shall call the displacement operator for the x-direction and denote by dx."

16. Jan 20, 2014

### Staff: Mentor

It does - and I apologize for the confusion - I replied without re-reading the book.

But it doesn't matter what you call it D displaces a state, dx is the derivative of D, and D goes to 1 as ∂x goes to zero.

Thanks
Bill

17. Jan 20, 2014

### strangerep

Just for the record, I'll note Dirac's eq(59)on p101:
$$|Pd\rangle ~=~ D|P\rangle ~,~~~~~~~ (59)$$and his words in the 2nd paragraph on p102 about how if we let $\delta x\to 0$ then "From physical continuity we should expect a displaced ket $|Pd\rangle$ to tend to the original $|P\rangle$". Hence $D\to 1$ in that limit.

All this is just application of Lie Group theory to physical space translations. The $D$ is a representation of a finite translation, and $d_x$ is the generator obtained by differentiating a general $D$ at the identity.

If this is the first time you're learning about Lie groups applied to physical spacetime symmetries, and their relevance in QM, I suggest you suspend reading Dirac and switch to Ballentine. It will be a lot easier to understand Dirac after you understand Ballentine.

18. Jan 21, 2014

### Staff: Mentor

I concur.

Dirac is a classic and should be in anyone's library with a serious interest in QM.

But it is NOT the best place to learn it. Ballentine is MUCH better.

I know this because I learned QM from Dirac - and it took me a while to unlearn some of its stuff such as exactly what a state is.

Thanks
Bill