If T unitary show H hermitian (and the reverse)

[ognik]

Homework Statement

An operator T(t+ε,t) describes the change in the wave function from to to t+ε. For ε real and small enough so that ε² may be neglected, considering the eqtn below:
(a)If T is unitary, show H is hermitian
(b)if H hermitian, show T is unitary.

Homework Equations

$$ T(t+ε,t)= 1-\frac{i}{\hbar} \epsilon H(t) $$

The Attempt at a Solution

I've been making decent progress through a bunch of hermitian/unitary exercises, but this looks nothing them - or anything in the text, which have been all about matrix operation without any functions... I think firstly there might be some ways to work with equations using U and H matrices, if so a link or 2 or a beginners explanation would be great. Then also a hint as to how I might start this? Thanks :-)

 

[robphy]

What are the definitions of being hermitian? unitary?

 

[ognik]

Thanks for the interest Robphy :-) I tried (a) If T unitary :
$$ T T^{\dagger} = 1 = (1-\frac{i}{\hbar}\epsilon H)(1-\frac{i}{\hbar}\epsilon H)^{\dagger} $$
$$ = (1-\frac{i}{\hbar}\epsilon H)(1+\frac{i}{\hbar}\epsilon H^{\dagger}) $$
But I've got something wrong? I can't work that so that $$ H=H^\dagger $$
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Trying (b), solve for H:
$$ H=-\frac{(T-1)}{\epsilon}\frac{\hbar}{i} = i(T-1)\frac{\hbar}{\epsilon}$$
If H hermitian, then
$$ H^\dagger =- i(T^\dagger-1)\frac{\hbar}{\epsilon}=i(T-1)\frac{\hbar}{\epsilon}$$
$$ \therefore -(T^\dagger-1)=(T-1)$$
which is also wrong. Must be in the way I found the hermitian of expressions, but I split that into conjugate first, then transpose - can't see what I'm doing wrong (after many pages of scrap paper). Please check my working and let me know what I've misunderstood?

 

[robphy]

For (a), carry out the multiplication.
Reread your problem statement.

For (b), I don't think it's a good idea to solve for H since (as you now hopefully saw in (a)) you need to use the approximation.
Assume the condition on H, what does it imply for T? Is the condition needed for T satisfied?

 

[ognik]

Thanks - good to know my arithmetic was ok, then for (a) I got
$$ 1-\frac{i}{\hbar} \epsilon H + \frac{i}{\hbar}\epsilon H^\dagger - \frac{-1}{\hbar^2} {\epsilon}^2HH^\dagger $$
Even though the problem says to ignore the ε² term, it would be multiplied by H²; also h bar is ~ 10^-34, so while I could ignore the ε² term, I can't comfortably see how to justify ignoring the whole of
$$ \frac{1}{\hbar^2} {\epsilon}^2HH^2 $$

I know the limit as ε→0 = 0, but (h bar)² is pretty small also; I have no feel for how small ε can be in practice... or how large H² can be?
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Similarly for (B), without solving for H - and having worked through all the hermitian properties I know of, the only thing that looked promising was to evaluate TT^*, which gave a similar expansion to the above. Then if H^*=H when H hermitian (which it doesn't) ... then TT^* also =1 if unitary ...
So, sorry if I'm being dense, but another hint at (b) please?
--------------
I have finished all the other exercises in this section, except another similar one - so excuse me for referencing 2 in 1, but its probably the same bit of understanding missing in both; so thought it might be useful for any helpers to know that ...
The other eqtn is U=exp(iaH), a real. I easily showed U unitary if H hermitian, but again can't see how to do the reverse (I definitely don't want to solve for H here ;-) ) Thanks again.

 

[Dick]

Thanks - good to know my arithmetic was ok, then for (a) I got
$$ 1-\frac{i}{\hbar} \epsilon H + \frac{i}{\hbar}\epsilon H^\dagger - \frac{-1}{\hbar^2} {\epsilon}^2HH^\dagger $$
Even though the problem says to ignore the ε² term, it would be multiplied by H²; also h bar is ~ 10^-34, so while I could ignore the ε² term, I can't comfortably see how to justify ignoring the whole of
$$ \frac{1}{\hbar^2} {\epsilon}^2HH^2 $$

I know the limit as ε→0 = 0, but (h bar)² is pretty small also; I have no feel for how small ε can be in practice... or how large H² can be?
---------
Similarly for (B), without solving for H - and having worked through all the hermitian properties I know of, the only thing that looked promising was to evaluate TT^*, which gave a similar expansion to the above. Then if H^*=H when H hermitian (which it doesn't) ... then TT^* also =1 if unitary ...
So, sorry if I'm being dense, but another hint at (b) please?
--------------
I have finished all the other exercises in this section, except another similar one - so excuse me for referencing 2 in 1, but its probably the same bit of understanding missing in both; so thought it might be useful for any helpers to know that ...
The other eqtn is U=exp(iaH), a real. I easily showed U unitary if H hermitian, but again can't see how to do the reverse (I definitely don't want to solve for H here ;-) ) Thanks again.

You shouldn't worry about ignoring terms like $\epsilon^2$, it's an infinitesimal. It's supposed to be much smaller than anything else around regardless of the other values (i.e. it's infinitesimal). Just ignore it. Then you've got, $TT^\dagger=1-\frac{i}{\hbar} \epsilon (H - H^\dagger)$. If $T$ is unitary then $TT^\dagger=1$, what does that tell you about $H-H^\dagger$? If $H$ is Hermitian what does that tell you about $TT^\dagger$ and $T$?

And it is true that showing $U=exp(iaH)$ is unitary implies $H$ is Hermitian is a bit trickier than the reverse. In particular, you don't know a priori that $H$ and $H^\dagger$ even commute. So be careful. Start from $UU^\dagger=1$ and differentiate both sides with respect to a. See where that leads you.

 

[Fredrik]

I don't think "we can ignore the $\varepsilon^2$ term because it's small" is the best argument here. The equality holds for all $\varepsilon$ in some small interval that contains 0. I would say that the statement that gives us the best reason to ignore the $\varepsilon^2$ isn't the "small", it's the "for all".

Consider the statement that a,b,c,r are real numbers such that $a+bx+cx^2=0$ for all real numbers x in the interval (-r,r). It implies that a=b=c=0. Similarly, if $a+bx+cx^2=d+ex+fx^2$ for all x, then the coefficients of each $x^n$ must be the same on both sides, so we have a=d, b=e, c=f.

If we apply this idea to the result that
$$1=1+\frac{i}{\hbar}\varepsilon(H^\dagger-H)+\frac{\varepsilon^2}{\hbar}H^\dagger H$$ for all $\varepsilon$ in some interval that contains 0, then the conclusion is that $H^\dagger=H$. At first it may seem that we can also conclude that $H^\dagger H=0$, but this would be wrong. The reason is the calculation has already neglected other terms of order $\varepsilon^2$, so the right-hand side is already wrong about second and higher order terms.

 

[ognik]

All done, many thanks - was much easier than it first seemed once I stopped being too impressed by the actual equations :-)

Could you also - perhaps & please - have a look at https://www.physicsforums.com/threa...te-systems-scale-factors.806607/#post-5064384 and see if you can shed some light? Its from a previous section of the book, but I prefer to finish everything - its bothering me that I got close but seem to be missing something in terms of index and summation...Thanks again.

 

[Fredrik]

There's something missing in your problem statement in that thread. You haven't defined the symbols and the notations you use. Am I right to assume that boldface symbols denote elements of $\mathbb R^3$ and that when you put a hat on such a symbol, it denotes the corresponding unit vector (e.g. $\hat{\mathbf q}=\mathbf q/|\mathbf q|$)? You're mixing your notations in a weird way. Is there a difference between $\hat{\mathbf q}_i$ and $\hat q_i$? Is the "relevant equation" supposed to involve $\mathbf r$ or $\hat{\mathbf r}$?

It's very likely that you didn't get a reply because no one understood the question. If you explain the problem statement (in the other thread), I can take another look at it.

 

[BvU]

Second Fredrik here. I had a look at that one too and couldn't make out heads or tails from the postings. A bit more context please.
My respect for your studying by correspondence and kudos for your tenacity. Keep it up and we'll help as best we can.

 

[ognik]

Thanks Frederiks :-) My last post must have crossed with yours, I appreciate the extra on the infinitesimals (and wouldn't have concluded H^†H =0). I have not really been happy with the way infinitesimals are sometimes used, I will continue to thing about them as I blunder on. Switching to other question now...

 

[Fredrik]

I think the best way to make sense of the term "infinitesimal" in a physics book is to interpret it as a code that let's the experienced reader know that the next calculation will involve a Taylor series expansion and that only terms up to some specific order will be included.

 

[ognik]

Thats kind of where I was, but it is sometimes unsatisfying ...The proof above doesn't work without it being 'neglected', so what does that say about T (and H) if we didn't have ε?

BTW, thanks for the encouragement BvU :-)

 

[Fredrik]

No, it does work. It's not about neglecting anything. It's about matching the coefficients of $\varepsilon^0,\varepsilon^1,\varepsilon^2,\dots$ on both sides. The fact that something had already been neglected in the calculation (actually in the definition of T) made it impossible to obtain a correct result from the coefficients of $\varepsilon^2$, but it had no effect on the coefficients on $\varepsilon^1$ or the coefficients of $\varepsilon^0$. We could neglect higher order terms because the result we're interested in is the equality of the coefficients of $\varepsilon$, but doing so had no advantages other than shortening the calculation.

 

[ognik]

that makes good sense thanks - we end up equating the coefficients of all sorts of things.

Any thoughts on the other problem after my latest post? Just let me know if you need anything else ...

 

[Fredrik]

I took a look at it yesterday. I think I understand all the notations and stuff, but I wasn't able to prove the desired result for $\frac{\partial\hat{\mathbf q}_i}{\partial q_j}$ when $i\neq j$. I will try again later today.