Differentiation of an exponential with operators (Peskin p.84)

In summary: Peskin has a problem with the exponential operator. He claims that it is difficult to differentiate between an exponential and a non-exponential property if there are several non-cummuting operators in the power of the exponential. He provides two formulas to illustrate this. The first formula shows that the operator is e^{iH_{0}\left( t-t_{0}\right) }, where H is the hamiltonian and t_{0} is the time at which the operator is evaluated. The second formula shows that the operator is e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }, where H is the hamiltonian and t is the time at
  • #1
gremezd
18
0
Does anyone know how to differentiate an exponential, which has an operator in its power? I found it quite a trouble in Peskin's QFT (page 84, formulas (4.17), (4.18)).
Here we have these two formulas of Peskin:

[tex]U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }e^{-iH\left( t-t_{0}\right) } [/tex];
[tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }\left( H-H_{0}\right) e^{-iH\left( t-t_{0}\right) } [/tex].

I agree with this. However, if we write [tex]U\left( t,t_{0}\right)[/tex] as [tex]U\left( t,t_{0}\right)=e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex], then

[tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=\left( H-H_{0}\right)e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex]

and we cannot transport [tex]e^{iH_{0}\left( t-t_{0}\right) }[/tex] to the left of [tex]\left( H-H_{0}\right)[/tex] so easily to obtain Peskin's result, since, according to my calculations, [tex]\left[ H,H_{0}\right]\neq0[/tex]. Do we have a rule, which explains where to put the operators from the exponential after differentiation, when we have several noncummuting operators in the power of exponential?
 
Physics news on Phys.org
  • #2
gremezd said:
Does anyone know how to differentiate an exponential, which has an operator in its power? I found it quite a trouble in Peskin's QFT (page 84, formulas (4.17), (4.18)).
Here we have these two formulas of Peskin:

[tex]U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }e^{-iH\left( t-t_{0}\right) } [/tex];
[tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }\left( H-H_{0}\right) e^{-iH\left( t-t_{0}\right) } [/tex].

I agree with this. However, if we write [tex]U\left( t,t_{0}\right)[/tex] as [tex]U\left( t,t_{0}\right)=e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex], then
But you can't write this since H and H_0 don't commute. [tex] e^A e^B = e^{A+B} [/tex] only when A and B commute. Otherwise you have to use the Campbell-Hausdorf formula.
[tex]i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=\left( H-H_{0}\right)e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }[/tex]

and we cannot transport [tex]e^{iH_{0}\left( t-t_{0}\right) }[/tex] to the left of [tex]\left( H-H_{0}\right)[/tex] so easily to obtain Peskin's result, since, according to my calculations, [tex]\left[ H,H_{0}\right]\neq0[/tex]. Do we have a rule, which explains where to put the operators from the exponential after differentiation, when we have several noncummuting operators in the power of exponential?
You just differentiate as usual, making sure that you never pass an operator "through" another operator that does not commute with it.
 
  • #3
Thanks a lot! This has been tormenting me for ages!
 
  • #4
Thank you, nrqed, for pointing out my mistake. I appreciate it :)
 
  • #5
gremezd said:
Thank you, nrqed, for pointing out my mistake. I appreciate it :)

:smile: You are very very welcome.

And thank you for posting your question since this apparently helped Wasia too!

Patrick
 

1. What is the general formula for differentiating an exponential with operators?

The general formula for differentiating an exponential with operators is d/dx (exp(a(x) * b(x))) = a'(x) * b(x) * exp(a(x) * b(x)) + a(x) * b'(x) * exp(a(x) * b(x)). This is known as the product rule for differentiation.

2. How is the product rule applied in differentiating an exponential with operators?

The product rule is applied by taking the derivative of the first function, multiplied by the second function, plus the first function multiplied by the derivative of the second function. This is then multiplied by the original exponential term.

3. Can the chain rule also be applied in differentiating an exponential with operators?

Yes, the chain rule can also be applied in differentiating an exponential with operators. This is done by taking the derivative of the outer function, and then multiplying it by the derivative of the inner function.

4. How do I differentiate an exponential with operators when both functions are dependent on the same variable?

If both functions are dependent on the same variable, you can use the quotient rule for differentiation. This is done by taking the derivative of the numerator multiplied by the denominator, minus the numerator multiplied by the derivative of the denominator. This is then divided by the square of the denominator.

5. Are there any special cases when differentiating an exponential with operators?

Yes, there are some special cases to keep in mind when differentiating an exponential with operators. These include when one or both functions are constants, or when one function is the natural logarithm. In these cases, the differentiation rules may be simplified.

Similar threads

  • Quantum Physics
Replies
10
Views
735
Replies
3
Views
1K
  • Quantum Physics
Replies
1
Views
629
  • Quantum Physics
Replies
1
Views
815
Replies
5
Views
844
Replies
3
Views
783
Replies
1
Views
770
Replies
2
Views
575
  • Quantum Physics
Replies
2
Views
690
Replies
3
Views
401
Back
Top