Dyson's Formula from Tong's lecture notes

[noir1993]

I am studying quantum field theory from [David Tong's lecture notes][1] and I am stuck at a particular place.

In Page 52., under the heading *3.1.1 Dyson's Formula*, Tong introduces an unitary operator
$$U(t, t_0) = T \exp(-i\int_{t_0}^{t}H_I(t') dt')$$

He then introduces the usual definition of time ordered products and goes on to expand $$U(t,t_0)$$. I am not able to follow how he expanded the time ordered product of operators in the second-order term of the Taylor expansion of the exponential. In particular, I am unable to follow the limits being used and why both integrals are being put in the front. Should we not get product of two integrals involving H_I?

The expansion of U(t,t_0) is given by

$$1 - i\int_{t_0}^{t}dt'H_I(t') + \frac{-i^2}{2}[\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t')+\int_{t_0}^{t}dt'\int_{t_0}^{t'}dt''H_I(t')H_I(t'')]+...$$

Link to Course Page - [David Tong: Lectures on Quantum Field Theory][2] [1]: http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf
[2]: http://www.damtp.cam.ac.uk/user/tong/qft.html

 

[vanhees71]

No, because there's the time-ordering symbol in front of the exponential function, which is defined by the series. For a different derivation of the same formula, see Sect. 1.3 in

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[hilbert2]

If you don't like the integration measures being in the front, you can use

$\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t') = \int_{t_0}^{t}\left(\int_{t'}^{t}H_I(t'')H_I(t')dt''\right)dt'$

The first (inner) integration converts $H_I(t'')H_I(t')$ to a function of $t'$ and $t$ only, and then the outer integration is over $t'$.

I personally find these calculations unpleasant too.

 

[noir1993]

If you don't like the integration measures being in the front, you can use

$\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t') = \int_{t_0}^{t}\left(\int_{t'}^{t}H_I(t'')H_I(t')dt''\right)dt'$

The first (inner) integration converts $H_I(t'')H_I(t')$ to a function of $t'$ and $t$ only, and then the outer integration is over $t'$.

I personally find these calculations unpleasant too.

It's not about the calculations. :D
It seems I just got confused by the notation. It makes sense now.

 

[formodular]

Hi!

Everytime I look at this I realize I have forgotten it.

How do you write up Tong's expansion so that it becomes clear?

$$T[\text{exp}(-i \int_{t_0}^t H_I(t')dt'] = T[1 + (-i) \int_{t_0}^t H_I(t')dt' + \frac{(-i)^2}{2!} \int_{t_0}^t dt'' \int_{t_0}^t dt' H_I(t')H_I(t'') + \dots]
= 1 + (-i) \int_{t_0}^t H_I(t')dt' + \frac{(-i)^2}{2!} T [\int_{t_0}^t dt'' \int_{t_0}^t dt' H_I(t')H_I(t'')] + \dots$$
?

 

[hilbert2]

Putting the differentials $dt''$, etc.. in front of the integrand is done just because it allows putting the integration "operator" $\int_{a}^{b}dt''$ in front of a function of $t''$ just like you can put a differential operator $\hat{p}_x = -i\hbar \frac{\partial}{\partial x}$ in front of a wave function $\psi (x)$. The only difference is that while $\hat{p}_x$ converts a function of $x$ to another function of $x$, the integral operator formally results in a function of a new variable.

 

[vanhees71]

The physicists' notation of integrals is very much more clever than the mathematians' who write the integration measure at the very end. I don't know of a better example than precisely the Dyson series.

The time-ordering is derived with some more steps in my QFT lecture notes (Sect. 1.3, p16ff):

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[noir1993]

With reference to comments made by @hilbert2 and @vanhees71, I have a question: If we treat the integration as operators then should not the time ordering now contain four operators instead of just two? Or am I missing something?

 

[vanhees71]

What do you mean by "four operators instead of just two"? The time-ordering symbol works at any product of an arbitrary number of field operators. It tells you to write the product such that the time arguments are increasing from the right- to the left-most operator. If the fields are fermionic it also includes the corresponding sign of the permutation needd to bring the operators to that order.

 

[noir1993]

The time-ordering is derived with some more steps in my QFT lecture notes (Sect. 1.3, p16ff):

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

It is sheer modesty to call this a set of lecture notes. It's basically a book. I have come across your notes before and have a copy saved in my reference manager. I decided to go through it after mastering the basics. In any case, I would like to thank you for making them available!

 

[vanhees71]

It's far from being a book! I'm most thankful if you find typos! ;-)

 

[noir1993]

What do you mean by "four operators instead of just two"? The time-ordering symbol works at any product of an arbitrary number of field operators. It tells you to write the product such that the time arguments are increasing from the right- to the left-most operator. If the fields are fermionic it also includes the corresponding sign of the permutation needd to bring the operators to that order.

I was talking with respect to the second order term again. I was under the impression that the operatiors being time ordered are the Hamiltonians H_I at different t's. But are we considering all the integrals ∫dt as operators also?

 

[vanhees71]

$\int \mathrm{d} t f(t)$ just tells you to do the integral.

 

[noir1993]

$\int \mathrm{d} t f(t)$ just tells you to do the integral.

I know that we can consider differentials and integrals as operators in the mathematical sense (as done in say, linear algebra or operator theory, i.e any transformation) but here I feel when we talk about time ordering of 'operators', we are talking about observables or rather "operators as understood in quantum theory".

 

[hilbert2]

Something like $H(t')H(t'')$ is an operator valued function of $t'$ and $t''$, where "operator" specifically means something that can act on a quantum state.

 

[noir1993]

Something like $H(t')H(t'')$ is an operator valued function of $t'$ and $t''$, where "operator" specifically means something that can act on a quantum state.

So this expression has four time ordered operators?

$\int_{t_0}^{t}dt'\int_{t'}^{t}dt''H_I(t'')H_I(t') = \int_{t_0}^{t}\left(\int_{t'}^{t}H_I(t'')H_I(t')dt''\right)dt'$

 

[hilbert2]

I don't think the integrations are same kind of operators as the $H_I$:s, as you can apply the integration on any function while the Hamiltonians only affect quantum states.

I similar ambiguous situation is how an $N\times N$ matrix can be seen either as a vector of dimension $N^2$ or an operator that acts from the left on any vector or matrix that has $N$ rows.

 

[noir1993]

I don't think the integrations are same kind of operators as the $H_I$:s, as you can apply the integration on any function while the Hamiltonians only affect quantum states.

A similar ambiguous situation is how an $N\times N$ matrix can be seen either as a vector of dimension $N^2$ or an operator that acts from the left on any vector or matrix that has $N$ rows.

Hence I posted a comment earlier stating this is in a different manner. I hope you can appreciate the reason behind my confusion. :)

 

[vanhees71]

So this expression has four time ordered operators?

I'm really puzzled, what you are talking about. There are only two operators in the expression, $\hat{H}_I(t'') \hat{H}(t')$. These you integrate
$$\hat{O}(t)=\int_{t_0}^t \mathrm{d} t' \int_{t'}^t \mathrm{d} t'' \hat{H}_I(t'') \hat{H}_I(t'). \qquad (1)$$
This is, of course another time-dependent operator.

The integral is obviously over a triangle in the $t'$-$t''$ plane, which is a bit inconvenient. The trick is to rewrite the integral such to integrate over $t'$ first:
$$\hat{O}(t)=\int_{t_0}^{t} \mathrm{d} t'' \int_{t_0}^{t''} \mathrm{d} t' \hat{H}_I(t'') \hat{H}_I(t').$$
Now we exchange the names of the integration variables, i.e., $t' \leftrightarrow t''$:
$$\hat{O}(t)=\int_{t_0}^t \mathrm{d} t' \int_{t_0}^{t'} \mathrm{d} t'' \hat{H}_I(t') \hat{H}_I(t''). \qquad (2)$$
Now we see that we could simply add this to (1), if the Hamiltonians at different times would commute, but in general they don't. However, we can write in both integrals a time-ordering symbol in front, because in both integrals the Hamiltonians are in time ordered form, i.e., the right-most operator has the smaller time argument. Since $\hat{H}_I$ must always be of bosonic nature (i.e., in the case of fermion fields there must be always an even number of field operators to build the Hamiltonian density) you can simply exchange the Hamiltonians as you like under the time-ordering operator, thus you can write for both integrands $\mathcal{T} \hat{H}_I(t'') \hat{H}_I(t')$ and then simply add the equations (1) and (2). Then you integrate over the entire square in the $t'$-$t''$ plane, and the integral looks a bit simpler:
$$2 \hat{O}(t)=\int_{t_0}^t \mathrm{d} t' \int_{t_0}^t \mathrm{d} t'' \mathcal{T} \hat{H}_I(t'') \hat{H}_I(t').$$
By induction you can show that this works also for the higher orders in the Dyson series (with an $n!$ on the left-hand side if you have $n$ factors $\hat{H}_I$ in the expression).

 

[noir1993]

I'm really puzzled, what you are talking about. There are only two operators in the expression, $\hat{H}_I(t'') \hat{H}(t')$. These you integrate
$$\hat{O}(t)=\int_{t_0}^t \mathrm{d} t' \int_{t'}^t \mathrm{d} t'' \hat{H}_I(t'') \hat{H}_I(t'). \qquad (1)$$
This is, of course another time-dependent operator.
.
.
.
.

Then you integrate over the entire square in the $t'$-$t''$ plane, and the integral looks a bit simpler:
$$2 \hat{O}(t)=\int_{t_0}^t \mathrm{d} t' \int_{t_0}^t \mathrm{d} t'' \mathcal{T} \hat{H}_I(t'') \hat{H}_I(t').$$
By induction you can show that this works also for the higher orders in the Dyson series (with an $n!$ on the left-hand side if you have $n$ factors $\hat{H}_I$ in the expression).

Thank you for this illuminating reply. It cleared every doubt that I had. I apologize for not being able to catch it early on from your comments and forced you to spell everything out! But I am very grateful for the answer!

And the digression about the "four operators" came from this comment where @hilbert2 called $\int_{a}^{b}dt''$ the integration "operator. It is not relevant to the main discussion at hand.

Putting the differentials $dt''$, etc.. in front of the integrand is done just because it allows putting the integration "operator" $\int_{a}^{b}dt''$ in front of a function of $t''$ just like you can put a differential operator $\hat{p}_x = -i\hbar \frac{\partial}{\partial x}$ in front of a wave function $\psi (x)$. The only difference is that while $\hat{p}_x$ converts a function of $x$ to another function of $x$, the integral operator formally results in a function of a new variable.