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Quantum Field Theory

  • #1
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I'm trying to understand path integrals as described in my lecture notes (which are reinforced by Peskin &Schroeder).

Anyway on p284 of P&S, there is a formula inbetween eqns (9.17) and (9.18) that reads:

[itex]e^{-iHT} | \phi_a \rangle = \sum_n e^{-i E_n T} | n \rangle \langle n | \phi_a \rangle \rightarrow \langle \Omega | \phi_a \rangle e^{-i E_0 \cdot \infty (1-i \epsilon)} | \Omega \rangle[/itex] as [itex]T \rightarrow \infty ( 1 - i \epsilon)[/itex]

I can follow the equality on the left fair enough but I don't understand what happens when we take the limit? Apparently, this is quite a common trick in QFT so can anybody explain to me what is going on here?

Thanks!
 

Answers and Replies

  • #2
fzero
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The presence of the [tex]\epsilon[/tex] means that the exponential is a strongly decaying function. The ground state term is slowest decaying term, so it dominates. You should refer to the discussion around equ. (4.27) for the relationship between the free and interacting ground states.
 
  • #3
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The presence of the [tex]\epsilon[/tex] means that the exponential is a strongly decaying function. The ground state term is slowest decaying term, so it dominates. You should refer to the discussion around equ. (4.27) for the relationship between the free and interacting ground states.
Ok. Thanks. Two things:

(i) In the expression I wrote down in my original post (out of my notes), why don't we set the [itex]e^{-E_o \infty (1-i \epsilon)}[/itex] factor to [itex]e^0=1[/itex]. Why do we leave it written out explicitly, even though we have taken the limit?

(ii) Just reading on in P&S, in (4.28), how do we go from the 2nd to the 3rd line?

Cheers!
 
  • #4
fzero
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Ok. Thanks. Two things:

(i) In the expression I wrote down in my original post (out of my notes), why don't we set the [itex]e^{-E_o \infty (1-i \epsilon)}[/itex] factor to [itex]e^0=1[/itex]. Why do we leave it written out explicitly, even though we have taken the limit?
In the scattering formalism, we treat the interaction as occuring in the time region around [tex]t=t_0[/tex]. At times far in the past and future, [tex]t\rightarrow\pm \infty[/tex], the particles are far enough from each other that they can be treated as free. The state [tex]|0\rangle[/tex] is the free vacuum, while [tex]|\Omega\rangle[/tex] is the vacuum for the interacting theory. On general principles, we know that

[tex] |\Omega\rangle = c_0 |0\rangle + O(T) + \cdots, (*)[/tex]

where [tex]c_0[/tex] is a c-number.

If we drop all [tex]T[/tex] dependence, we lose all information about anything past the first term in (*).

(ii) Just reading on in P&S, in (4.28), how do we go from the 2nd to the 3rd line?

Cheers!
Use the definition of [tex]U[/tex] from (4.17).
 
  • #5
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In the scattering formalism, we treat the interaction as occuring in the time region around [tex]t=t_0[/tex]. At times far in the past and future, [tex]t\rightarrow\pm \infty[/tex], the particles are far enough from each other that they can be treated as free. The state [tex]|0\rangle[/tex] is the free vacuum, while [tex]|\Omega\rangle[/tex] is the vacuum for the interacting theory. On general principles, we know that

[tex] |\Omega\rangle = c_0 |0\rangle + O(T) + \cdots, (*)[/tex]

where [tex]c_0[/tex] is a c-number.

If we drop all [tex]T[/tex] dependence, we lose all information about anything past the first term in (*).



Use the definition of [tex]U[/tex] from (4.17).
I now get the first bit.

However, I have in my notes that these time evolution operators are given by Dyson's formula:

[itex]U(t,t_0)= Te^{-i \int_{t_0}^t \hat{H}(t') dt'}[/itex]

Are these equivalent?
 
  • #6
fzero
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Dyson's formula is shown to be equivalent to (4.17) in the discussion following that formula.
 
  • #7
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Dyson's formula is shown to be equivalent to (4.17) in the discussion following that formula.
Ok. I'm not sure how easy this next bit will be for you to follow as I haven't been able to find t he corresponding stuff in P&S so this is all just from my notes.

I have the formula:

[itex]Z ( \vec{b})=\frac{1}{Z_A} \int d^Nx e^{-\frac{1}{2} \vec{x}^T \cdot A \vec{x} + \vec{b}^T \cdot \vec{x} - V( \vec{x} ) }[/itex] (*)

where [itex]Z_A = \int d^N x e^{-\frac{1}{2} \vec{x}^T \cdot A \vec{x}}[/itex]

and [itex]Z_{A, \vec{b}} = \int d^N x e^{-\frac{1}{2} \vec{x}^T \cdot A \vec{x} + \vec{b}^T \cdot \vec{x}}[/itex]

Now we are told that (*) converges if V is bounded below and that for simplicity, [itex]V(0)=0, \frav{\partial V}{\partial \vec{x}} |_{\vec{x}=0}=0[/itex]

Now I need to figure out how to write this as

[itex]Z( \vec{b})=e^{-V( \frac{\partial}{\partial \vec{b}} )} \frac{1}{Z_A} Z_{A, \vec{b}}[/itex]

where [itex]V( \vec{b} ) = \sum_{n=0}^\infty \frac{1}{n!} V^{(n)}_{i_1, \dots , i_n}(0) \frac{\partial}{\partial b_i_1} \dots \frac{\partial}{\partial b_i_n}[/itex]

Hopefully this makes sense or you've seen something like this before!

Thanks.
 
  • #8
fzero
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What have you tried so far? That doesn't seem like it's difficult at all to verify.
 
  • #9
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What have you tried so far? That doesn't seem like it's difficult at all to verify.
Yeah I managed it today by starting with the answer and working backwards....which is probably what I should have been doing in the first place!!!

I do have another question though:

if [itex]K(q,q_0;t-t_0)= \left( \frac{m}{2 \pi i (t-t_0)} \right)^{ \frac{1}{2}} e^{i m \frac{(q-q_0)^2}{2(t-t_0)}}[/itex] with [itex]t>t_0[/itex]

I was asked to show that [itex]-\frac{1}{2m} \frac{\partial^2}{\partial x^2} K(x,0;t) = i \frac{\partial}{\partial x} K(x,0;t)[/itex]
I managed that fine - it was just a simple exercise in differentiation.

Now I have been asked to express the solutions of the Schrodinger equation for a free particle in terms of initial data [itex]\psi(x,0)[/itex] and [itex]K(x,0;t)[/itex].

So can I write [itex]\psi(x,t)=K(x,0;t) \psi(x,0)[/itex]? since K describes the amplitude for a free particle to propagate from [itex](x_0,t_0)[/itex] to [itex](x,t)[/itex]. I would have thought that using the definition of K that I just mentioned, we should be writing [itex]\psi(x,t)=K(x,0;t) \psi(0,0)[/itex] since our [itex]x_0=0[/itex], no?

Anyway, regardless of that, I'm not sure what it wants me to do with this? Does it just want me to write

[itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} (K(x,0;t) \psi(x,0))=i \hbar \frac{\partial}{\partial t} (K(x,0;t) \psi(x,0))[/itex]

I assume I'm meant to do this differentiation now? Essentially this gives

[itex]-\frac{\hbar^2}{2m} (K' \psi+K \psi')' = i \hbar (\dot{K} \psi + K \dot{\psi}) \Rightarrow -\frac{\hbar^2}{2m} (K'' \psi + 2K' \psi' + K \psi'') = i \hbar (\dot{K} \psi + K \dot{\psi})[/itex]
But then the [itex]K''[/itex] term will cancel the [itex]\dot{K}[/itex] term, right? Leaving

[itex]-\frac{\hbar^2}{2m} (2K' \psi' + K \psi'') = i \hbar ( K \dot{\psi})[/itex]

Now should I also get rid of the [itex]\dot{\psi}[/itex]? I guess we are talking about [itex]\dot{\psi(x,0)}[/itex] here so it will not have any time dependence, right? So I think it should go as well leaving

[itex]-\frac{\hbar^2}{2m} (2K' \psi' + K \psi'')[/itex]
[itex]2K' \psi' + K \psi''=0[/itex]

How does that look? I'm fairly sure there's an error because we are then asked to check for [itex]\psi(x,0)=e^{ikx}[/itex] but the [itex] \psi'[/itex] term will only have k in it and the [itex] \psi''[/itex] term will have a k^2 so they won't be able to cancel, will they?

Thanks.
 
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  • #10
fzero
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What's true is that

[itex]
\psi(x,t)=\int d^3x' K(x,x';t) \psi(x',0),
[/itex]

so I would actually start from that. This satisfies Schrodinger's equation, but it doesn't appear to reduce to the formula you wrote down in any way.
 
  • #11
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yeah its actually

[tex]\psi(x,t) = \int K(x-y,t)\psi(y,0) dy[/tex] which satisfies all the requirements such as the Schroedinger equation and [tex] \lim_{t \to 0}\psi(x,t) =\psi(x) [/tex] the question is a bit misleading in that sense
 
  • #12
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What's true is that

[itex]
\psi(x,t)=\int d^3x' K(x,x';t) \psi(x',0),
[/itex]
doesn't this make more sense

[itex]\psi(x,t)=\int d^3x' K(x,x';t) \psi(x',t),[/itex]

since

[itex]K(x,x';t) = \left\langle \psi(x,t)\right| e^{-iHt} \left|\psi(x',0)\right\rangle[/itex] i.e. the coefficients necessary to express [tex]\psi(x,t)[/tex] in terms of [tex]\psi(x',t)[/tex]

however they do both satisfy

[itex] \lim_{t \to 0} \psi(x,t) = \psi(x,0) [/itex]

and i'm guessing that your one is more attractive since it easily obeys the Schroedinger equation
 
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  • #13
fzero
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doesn't this make more sense

[itex]\psi(x,t)=\int d^3x' K(x,x';t) \psi(x',t),[/itex]
Not really. Using the notation here,

[itex]\psi(x,t)=\int d^3x' K(x,x';t-t_0) \psi(x',t_0).[/itex]

You certainly don't propagate a state at time [tex]t[/tex] to another at time [tex]t[/tex].
 
  • #14
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K is not an evolution operator its an amplitude. Anyway [tex]\psi(x',0)[/tex] is a solution to the Schroedinger equation so it doesn't matter.
 
  • #15
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Not really. Using the notation here,

[itex]\psi(x,t)=\int d^3x' K(x,x';t-t_0) \psi(x',t_0).[/itex]

You certainly don't propagate a state at time [tex]t[/tex] to another at time [tex]t[/tex].
Ok.So,

[itex]\psi(x,t) = \int d^3x' K(x,x';t) \psi(x',0)[/itex]

So the Schrodinger equation implies
[itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \int d^3 x' K(x,x';t) \psi(x',0) = i \hbar \frac{\partial}{\partial t} \int d^3x' K(x,x';t) \psi(x',0)[/itex]
[itex]- \frac{\hbar^2}{2m} \int d^3x' \frac{\partial^2 K}{\partial x^2} \psi + 2 \frac{\partial K}{\partial x} \frac{\partial \psi}{\partial x} + K \frac{\partial^2 \psi}{\partial x^2} = i \hbar \int d^3 x' \dot{K} \psi + K \dot{\psi}[/itex]
[itex]- \frac{\hbar^2}{2m} \int d^3x' \frac{\partial^2 K}{\partial x^2} \psi + 2 \frac{\partial K}{\partial x} \frac{\partial \psi}{\partial x} + K \frac{\partial^2 \psi}{\partial x^2} = i \hbar \int d^3 x' \dot{K} \psi [/itex] as [itex]\dot{\psi}=0[/itex]

But then I don't know how to simplify this from here? Is this as simple as it gets? The final part is to check this result for [itex]\psi(x,0)=e^{ikx}[/itex], so should I just substitute that in and see if the LHS=RHS?

Also, how would we show that [itex] \lim_{t \rightarrow 0} K(x,0;t) = \delta(x)[/itex]

Thanks very much!
 
  • #16
fzero
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Ok.So,

[itex]\psi(x,t) = \int d^3x' K(x,x';t) \psi(x',0)[/itex]

So the Schrodinger equation implies
[itex]-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \int d^3 x' K(x,x';t) \psi(x',0) = i \hbar \frac{\partial}{\partial t} \int d^3x' K(x,x';t) \psi(x',0)[/itex]
[itex]- \frac{\hbar^2}{2m} \int d^3x' \frac{\partial^2 K}{\partial x^2} \psi + 2 \frac{\partial K}{\partial x} \frac{\partial \psi}{\partial x} + K \frac{\partial^2 \psi}{\partial x^2} = i \hbar \int d^3 x' \dot{K} \psi + K \dot{\psi}[/itex]
[itex]- \frac{\hbar^2}{2m} \int d^3x' \frac{\partial^2 K}{\partial x^2} \psi + 2 \frac{\partial K}{\partial x} \frac{\partial \psi}{\partial x} + K \frac{\partial^2 \psi}{\partial x^2} = i \hbar \int d^3 x' \dot{K} \psi [/itex] as [itex]\dot{\psi}=0[/itex]
Be more careful, [tex]\partial \psi(x',0) /\partial x = 0[/tex].


But then I don't know how to simplify this from here? Is this as simple as it gets? The final part is to check this result for [itex]\psi(x,0)=e^{ikx}[/itex], so should I just substitute that in and see if the LHS=RHS?
Schrodinger's equation is satisfied from the result from post 9:

[itex]
-\frac{1}{2m} \frac{\partial^2}{\partial x^2} K(x,0;t) = i \frac{\partial}{\partial x} K(x,0;t)
[/itex]

The same equation holds for [tex]x'\neq 0[/tex].

Also, how would we show that [itex] \lim_{t \rightarrow 0} K(x,0;t) = \delta(x)[/itex]

Thanks very much!
You should probably look at a sketch of the exponential expression to see what's happening around [tex]x=0[/tex].
 
  • #17
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Be more careful, [tex]\partial \psi(x',0) /\partial x = 0[/tex].
So getting rid of those terms leaves
[itex]\frac{\hbar^2}{2m} \frac{\partial}{\partial x} \int d^3x' \frac{\partial K(x,x';t)}{\partial x} \psi(x',0) = i \hbar \int d^3 x' \dot{K}(x,x';t) \psi(x',0)[/itex]
[itex]-\frac{\hbar^2}{2m} \int d^3 x' \frac{\partial^2 K(x,x';t)}{\partial x^2} \psi(x',0) = i \hbar \int d^3x' \dot{K}(x,x';t) \psi(x',0)[/itex]

Schrodinger's equation is satisfied from the result from post 9:

[itex]
-\frac{1}{2m} \frac{\partial^2}{\partial x^2} K(x,0;t) = i \frac{\partial}{\partial x} K(x,0;t)
[/itex]

The same equation holds for [tex]x'\neq 0[/tex].
Ok. Well I don't see how this can be applied to what I wrote above so I must have got the above wrong I think.


You should probably look at a sketch of the exponential expression to see what's happening around [tex]x=0[/tex].
Well [itex]K(x,0;t)=\left( \frac{m}{2 \pi i t} \right)^2 e^{\frac{imx^2}{2t}}[/itex]

So as [itex]t \rightarrow 0[/itex], the coefficient will blow up to infinity but, more importantly (as it will blow up much faster), the exponential tends to [itex]e^\infty[/itex] at teh origin thus giving a delta function. Is this question answered just by arguing about behaviour rather than actually doing any computations?
 
  • #18
fzero
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So getting rid of those terms leaves
[itex]\frac{\hbar^2}{2m} \frac{\partial}{\partial x} \int d^3x' \frac{\partial K(x,x';t)}{\partial x} \psi(x',0) = i \hbar \int d^3 x' \dot{K}(x,x';t) \psi(x',0)[/itex]
[itex]-\frac{\hbar^2}{2m} \int d^3 x' \frac{\partial^2 K(x,x';t)}{\partial x^2} \psi(x',0) = i \hbar \int d^3x' \dot{K}(x,x';t) \psi(x',0)[/itex]



Ok. Well I don't see how this can be applied to what I wrote above so I must have got the above wrong I think.
Well you had a typo in that formula, the RHS should be a time derivative.

As far as the above formulas, you shouldn't be writing down a RHS and LHS and trying to manipulate both. Start from one side and attempt to show the other, so you'd have

[itex]\frac{\hbar^2}{2m} \frac{\partial}{\partial x} \int d^3x' \frac{\partial K(x,x';t)}{\partial x} \psi(x',0) =-\frac{\hbar^2}{2m} \int d^3 x' \frac{\partial^2 K(x,x';t)}{\partial x^2} \psi(x',0) .[/itex]

Now use the earlier result that you derived to relate this to the time derivative and hence show that Schrodinger's equation is satisfied.

Well [itex]K(x,0;t)=\left( \frac{m}{2 \pi i t} \right)^2 e^{\frac{imx^2}{2t}}[/itex]

So as [itex]t \rightarrow 0[/itex], the coefficient will blow up to infinity but, more importantly (as it will blow up much faster), the exponential tends to [itex]e^\infty[/itex] at teh origin thus giving a delta function. Is this question answered just by arguing about behaviour rather than actually doing any computations?
To prove it more formally, you'd have to use a suitable integral expression to verify the delta function behavior. However, this is a reasonably well-known representation of the delta function. There are also other expressions for the propagator that are better suited to verify delta functions in limits (Sakurai is one place to find the discussion).
 
  • #19
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Well you had a typo in that formula, the RHS should be a time derivative.

As far as the above formulas, you shouldn't be writing down a RHS and LHS and trying to manipulate both. Start from one side and attempt to show the other, so you'd have

[itex]\frac{\hbar^2}{2m} \frac{\partial}{\partial x} \int d^3x' \frac{\partial K(x,x';t)}{\partial x} \psi(x',0) =-\frac{\hbar^2}{2m} \int d^3 x' \frac{\partial^2 K(x,x';t)}{\partial x^2} \psi(x',0) .[/itex]

Now use the earlier result that you derived to relate this to the time derivative and hence show that Schrodinger's equation is satisfied.



To prove it more formally, you'd have to use a suitable integral expression to verify the delta function behavior. However, this is a reasonably well-known representation of the delta function. There are also other expressions for the propagator that are better suited to verify delta functions in limits (Sakurai is one place to find the discussion).
Ok. Think I get it now.

One little thing though.

You said that [itex]\psi(x,t) = \int d^3 x' K(x,x';t) \psi(x',0)[/itex]

That's fair enough, we're integrating over all possible positions [itex]x'[/itex] where the particle could ahve started from, right?

But, as in this case, if I am asked to express [itex]\psi(x,t)[/itex] in terms of initial data [itex]K(x,0;t)[/itex] and [itex]\psi(x,0)[/itex]

Then surely either there is a misprint and the initial data should be [itex] \psi(0,0)[/itex]? And in that case, we would have fixed the initial position so there would be no need to integrate it and so we would have [itex]\psi(x,t)=K(x,0;t) \psi(0,0)[/itex]

And for proving the schrodigner situation holds, i started with the LHS (with the [itex]\hbar^2[/itex]) but was unable to get rid of one fo the factors of [itex]\hbar[/itex] - we expect teh RHS to be [itex]i \hbar \frac{\partial}{\partial t} \psi(x,t)[/itex] but I'm out by this factor - any advice?

Thanks again,
 
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  • #20
fzero
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Ok. Think I get it now.

One little thing though.

You said that [itex]\psi(x,t) = \int d^3 x' K(x,x';t) \psi(x',0)[/itex]

That's fair enough, we're integrating over all possible positions [itex]x'[/itex] where the particle could ahve started from, right?

But, as in this case, if I am asked to express [itex]\psi(x,t)[/itex] in terms of initial data [itex]K(x,0;t)[/itex] and [itex]\psi(x,0)[/itex]

Then surely either there is a misprint and the initial data should be [itex] \psi(0,0)[/itex]? And in that case, we would have fixed the initial position so there would be no need to integrate it and so we would have [itex]\psi(x,t)=K(x,0;t) \psi(0,0)[/itex]
I'd assume that it's a misprint, because [itex]\psi(x,t)=K(x,0;t) \psi(0,0)[/itex] makes no sense. You should go back and review the definition of the propagator in terms of quantum states to see this.

And for proving the schrodigner situation holds, i started with the LHS (with the [itex]\hbar^2[/itex]) but was unable to get rid of one fo the factors of [itex]\hbar[/itex] - we expect teh RHS to be [itex]i \hbar \frac{\partial}{\partial t} \psi(x,t)[/itex] but I'm out by this factor - any advice?

Thanks again,
You're missing some factors of [tex]\hbar[/tex] in your expression for the propagator. The proper expression should be

[itex]
K(q,q_0;t-t_0)= \left( \frac{m}{2 \pi i \hbar (t-t_0)} \right)^{ \frac{1}{2}} e^{i m \frac{(q-q_0)^2}{2\hbar(t-t_0)}}
[/itex]
 
  • #21
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you should really read the question sheet because it says at the top that hbar =1 throughout

and I've already given you the solution

[tex]\psi(x,t) = \int K(x-x',0;t)\psi(x',0) dx'[/tex]
 
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  • #22
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you should really read the question sheet because it says at the top that hbar =1 throughout

and I've already given you the solution

[tex]\psi(x,t) = \int K(x-x',0;t)\psi(x',0) dx'[/tex]
For the last bit where we are asked to check this for [itex]\psi(x,0)=e^{ikx}[/itex]

So we are asked here to find [itex]\psi(x,t)[/itex] I guess.

So [itex]\psi(x,t) = \int d^3 x' \left( \frac{m}{2 \pi i t} \right)^{\frac{1}{2}} e^{\frac{im(x-x')^2}{2t}} e^{ikx'}[/itex]
[itex]\psi(x,t) = \left( \frac{m}{2 \pi i t} \right)^{\frac{1}{2}} e^{\frac{imx^2}{2t}} \int d^3 x' e^{\frac{im}{2t} ( x'^2 - 2xx' + \frac{2kt}{m}x')}[/itex]

But I don't know how to do that integral?
 
  • #23
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you complete the square

if you want the solutions are up on the web

http://www.damtp.cam.ac.uk/user/me288/teaching/aqft/aqft_solutions_1.pdf [Broken]
 
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  • #24
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you complete the square

if you want the solutions are up on the web

http://www.damtp.cam.ac.uk/user/me288/teaching/aqft/aqft_solutions_1.pdf [Broken]
In the second last line, how does he get

[itex]\int d^y e^{\frac{im}{2t} (x'+\frac{kt}{m}-x)^2} = \int dy e^{-\frac{m}{2it}y^2}[/itex]

what happened to the 2nd and 3rd terms?
 
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  • #25
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he used a change of variable, but he never bothered to change the notation. A clearer way to write what he did would be

[itex]
\int dy e^{\frac{im}{2t} (y+\frac{kt}{m}-x)^2} = \int dy' e^{-\frac{m}{2it}y'^2}
[/itex]

where [tex] y' =y+\frac{kt}{m}-x[/tex] from which it is clear that [tex]\frac{dy'}{dy}=1[/tex]
 

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