# What is the effect of e^(isH)? p.77, Ballentine

1. Apr 27, 2012

### syhpehtam

Hi guys, I am reading Ballentine's Quantum Mechanics book, and I am stuck on p.77.
I don't know how the above formula comes, what is the effect of $e^{isH}$ when it acts on a vector?
In this page, there is also a sentence that I don't understand:
Does this sentence imply something that can derive the above formula?

2. Apr 27, 2012

### Fredrik

Staff Emeritus
$|\Psi(t)\rangle$ is the vector that you would use to represent the state of the system at time t. $|\Psi(t-s)\rangle$ is the vector you would use to represent the state of the same system s seconds earlier, or equivalently, the vector you would have used instead of $|\Psi(t)\rangle$ if you had been using a coordinate system in which the time coordinate reaches 0 s seconds later than in the coordinate system you're using now.

We can define a family of unitary operators (one for each s) by $U(s)|\Psi(t)\rangle=|\Psi(t+s)\rangle$. We assume that the function U satisfies U(t+s)=U(t)U(s). This is how the idea of translational invariance of time is incorporated into QM. Now there's a theorem (Stone's) that tells us that since U has this property, there's a unique self-adjoint operator H such that U(t)=exp(-iHt) for all t. The Hamiltonian is defined as that operator.

The function U in this case is a group homomorphism from the group $(\mathbb R,+)$ into the group of unitary operators on our Hilbert space. The group $(\mathbb R,+)$ is (isomorphic to) the group of translations in time. To claim that "the dynamics of a free particle are invariant under the full Galilei group of space-time transformations" is to claim that we can do essentially the same thing with any one-parameter subgroup of the Galilei group, not just the time translation group.

3. Apr 27, 2012

### strangerep

The action of transformations of wavefunctions was already explained in sections 3.1 and 3.2. The details of the Galilei group of transformations was explained in sections 3.2 & 3.3. So I'm kinda surprised that you had trouble on p77.

Maybe you should re-study those earlier sections and see whether there's still anything in there you don't understand.

No. Up to this point he's been using the familiar properties of space and time to derive stuff. E.g., he derived earlier a geometric quantity "P" that generates spatial translations. But to make the leap from there to seeing "P" as being also the familiar dynamical quantity we call "momentum", an extra step is necessary -- which is what section 3.4 is all about.

4. May 18, 2012

### syhpehtam

I am sorry for replying so late. It seems that Ballentine uses $|\Psi(t)\rangle\rightarrow e^{isH}|\Psi(t)\rangle=|\Psi(t-s)\rangle$ to define the operator $H$, the formula is a definition. Are formulas (3.40)$|\mathbf x\rangle\rightarrow|\mathbf x\rangle'=e^{-i\mathbf a\cdot\mathbf P}|\mathbf x\rangle=|\mathbf x+\mathbf a\rangle$, (3.45)$|\mathbf x\rangle\rightarrow|\mathbf x\rangle'=e^{-i\theta\hat{n}\cdot\mathbf J}|\mathbf x\rangle=|\mathbf x'\rangle$ and (3.49)$e^{i\mathbf v\cdot\mathbf G}\mathbf Ve^{-i\mathbf v\cdot\mathbf G}=\mathbf V-\mathbf vI$ also definitions which define operators $\mathbf P$, $\mathbf J$ and $\mathbf G$? If so, why those operators should be defined like this? And how can Ballentine know that those opertors which he defined satisfy the commutation relations(3.35)? I feel that I really don't know what section 3.4 is talking about.

I think that section 3.1~3.3 is talking about some mappings $U(\tau)$, these mappings satisfy the following conditions:
1)the value $|\langle\phi|\psi\rangle|$ should be preserved.
2)$U(\tau_2\tau_1)=e^{i\omega}U(\tau_2)U(\tau_1)$, and when $\tau_1$,$\tau_2$ are both the same kind of elementary transformations like rotations, displacements, etc., $\omega=0$.

We know that for any set of mappings satisfying the above conditions, there correspond a set of operators, namely $\mathbf J$,$\mathbf P$,$\mathbf G$ and $H$. And we can choose some special sets of mappings so that the commutation relations between their opertors can be simplified to formula(3.35).

This is what I've got from those sections, I don't know whether or not I misunderstood something.

Last edited: May 18, 2012
5. May 18, 2012

### syhpehtam

From the last two paragraphs of sec.3.2, does the author regard the state vector $|\Psi\rangle$ as a function $\Psi:(x,y,z,t)\mapsto\Psi(x,y,z,t)$, and $U(\tau)|\Psi\rangle=|\Psi'\rangle$ such that $\Psi'(x,y,z,t)=e^{i\omega}\Psi(\tau^{-1}(x,y,z,t))$?
The set of mappings $U(\tau)$ satisfy the conditions in #4.

Last edited: May 18, 2012
6. May 21, 2012

### strangerep

They are defined as generators of the Galilei group in section 3.3 (which assumes the reader has understood sect 3.2). This is crucial. You won't understand later sections unless you understand the concept of a Lie group and its infinitesimal generators, and the applicability of these to spacetime transformations.

After that, the next step is to understand the concept of representing these abstract quantities as operators on a Hilbert space of abstract states, and then on a Hilbert space of wavefunctions. E.g., $|\Psi\rangle$ is an abstract state in an abstract Hilbert space, whereas $\Psi(x)$ is a wavefunction in a Hilbert space of square-integrable functions.
This is what Ballentine means in the 1st sentence of the last paragraph on p67 where he
says "[...] when the abstract vector $|\Psi\rangle$ is represented as a function of space-time coordinates [...]".

7. May 21, 2012

### syhpehtam

I haven't learnt Lie group before. I only know a little elementary knowledge about groups. I think I should learn the mathematics of Lie group or group theory which you mentioned. Can you please recommend some books from which I can learn this kind of mathematics? I've found the book "Lie Algebras in Particle Physics" by Howard Georgi, is this book suitable?

Last edited: May 21, 2012
8. May 21, 2012

### dm4b

Are the exponents throwing you off? If so, you just need to understand how to go from infinitesimal to finite transformations.

I don't think you should need to read a whole book on Lie Algebra to understand that. Some QM books have a decent enough explanation. I prefer Shankar, myself, and he devotes at least a section, or two, to showing this.

9. May 21, 2012

### syhpehtam

I don't need to read the whole book, I can choose a few chapters on the mathematics which I concern about to read.
I know Shankar is very good, but I don't know if it is as advanced as Sakurai or Ballentine. Is Shankar enough before proceeding to QFT?

10. May 21, 2012

### dm4b

It was for me. Although, I did use David Tong's QFT text/manual as a baby step before starting out on Peskin and Schroeder.

11. May 21, 2012

### strangerep

I haven't read Georgi's book, so I can't give an opinion.
But I'm sure this book would be suitable:

Greiner & Muller, "Quantum Mechanics -- Symmetries".

12. May 22, 2012

### syhpehtam

Thank you. I'll try the book you recommended.