# Path Integral Propagator Normalization in Lewis Ryder's QFT book

1. Aug 9, 2009

### maverick280857

Hi,

In Lewis Ryder's QFT book on page 160, the propagator for the case when the Lagrangian can be written as

$$L = \frac{p^2}{2m} + V(q)$$

is given as

$$\langle q_f t_f|q_i t_i \rangle = \lim_{n\rightarrow\infty}\left(\frac{m}{i\hbar\tau}\right)^{(n+1)/2}\int \Pi_{1}^{n}dq_{j}\exp{\left\{\frac{i\tau}{\hbar}\sum_{0}^{n}\left[\frac{m}{2}\left(\frac{q_{j+1}-q_{j}}{2}\right)^{2}-V\right]\right\}}$$

and hence in the "continuum limit" as

$$\langle q_f t_f|q_i t_i \rangle = N\int\mathcal{D}q\exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}$$

To quote the author,

Now, I'm guessing $N = \left(\frac{m}{i\hbar\tau}\right)^{(n+1)/2}$. Am I right? Also, what does he mean by "we shall always deal with the normalized transition amplitudes"? What is the normalization here?

PS -- How does one get the nice looking caligraphic D for the functional integral? I used \mathcal{D}, but the effect isn't as pronounced as in Ryder's book

2. Aug 9, 2009

### RedX

The 2 in the denominator inside the squared-velocity should probably be a tau, and the Lagrangian should probably have minus the potential in it, and not positive the potential.

You're right about 'N'. It goes to infinity as tau, the discrete time interval, goes to zero. Normalization means that $$| \langle q_f t_f|q_i t_i \rangle |^2$$
summed over all final states $$|q_f,t_f>$$ equals one, since at time $$t_f$$ the particle has to be in some $$q_f$$. So the N doesn't really matter since you multiply every amplitude by the same constant factor to get the total probability of being somewhere add up to one.

3. Aug 10, 2009

### maverick280857

Yes, you're right. Typographical mistake, sorry about that.

Ok, but then

$$|\langle q_f t_f|q_i t_i\rangle|^2$$

is proportional to $N^2$, and if you sum it over all possible final states $|q_f t_f\rangle$, you will get a sum multiplied by $N^2$. If this sum is to become one even as $N \rightarrow \infty$ shouldn't the sum it multiplies tend to zero or something?

I am using

$$\langle q_f t_f|q_i t_i\rangle = N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}$$

to make this assertion.

4. Aug 10, 2009

### RedX

I'm not entirely sure I understand how something as simple as $$\langle q_f t_f|q_i t_i\rangle$$
happens to be infinity. My guess is that in order to derive the path integral, one inserts too many intermediate states in between the initial and final states (using the completeness relations) and that's where the infinity comes from.

Nevertheless, I would view this normalization as similar to playing the lottery. The more times you play, the more times you win. However, this should not affect the probability. So if you win the lottery 10 times, you have to divide by the 100 times you played, for a 10% winning probability. Similarly, I would view what you're calculating as not the probability, but how many times you've won, or how many times the particle is found at $$q_f$$. So what you have is relative probabilities - you can tell that one state is 3 times more likely if you win with that state 3 times more. To get the absolute probability, you have to divide by an overall constant to get everything to add up to 1.

5. Aug 11, 2009

### maverick280857

Ok, my first thought was that in normalizing the amplitude, the N is cancelled.

Anyway since the author claims there is no problem, N is really going away right?

(PS -- Sorry if this is a trivial question, but I couldn't see how the normalization worked.)

6. Aug 12, 2009

### maverick280857

Any ideas?

7. Aug 12, 2009

### jensa

Well, I'm not sure I can answer your question the way you would like it but I think you should always think of the right hand side of

$$\langle q_f t_f|q_i t_i\rangle = N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}$$

as a "short hand notation" for the discretized form. In the case of a free particle ($V=0$) you can check by explicit calculation of the discretized form that it is finite and indeed gives you the correct form of the propagator using $(n+1)\tau=t_f-t_i$. When evaluating path integrals for $V\neq 0$, one usually uses the free propagator as a "comparison"/normalization to obtain the correct form. For example:

$$\langle q_f t_f|q_i t_i\rangle = N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}=\frac{N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}}{\langle q_f t_f|q_i t_i\rangle_0}\langle q_f t_f|q_i t_i\rangle_0$$

where $\langle q_f t_f|q_i t_i\rangle_0$ denotes the propagator of a free particle. Then inserting in the denominator the path integral expression for the free particle you find that the "normaliziation factor" $N$ cancels.

$$\langle q_f t_f|q_i t_i\rangle = N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}=\frac{ \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}}{\int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L^0(q,\dot{q}) dt\right]}}\langle q_f t_f|q_i t_i\rangle_0$$

where $L^0=m\dot{q}^2/2$, i.e. the Lagrangian of a free particle. Then finally you can replace $\langle q_f t_f|q_i t_i\rangle_0$ with the explicit form of the free particle propagator. This way we don't have to deal with the factor $N$.

Last edited: Aug 12, 2009
8. Aug 14, 2009

### maverick280857

Thanks jensa and RedX, I think I have a better understanding of this now. I'll use this thread if I have any more queries regarding the normalization.

9. Aug 14, 2009

### jensa

Maverick,

Just wanted to add something on top of what I said before. Since almost all we can do with path integrals boils down to some combination of perturbation expansion, saddlepoint approximations and gaussian integrals it is useful to look at the case with quadratic potential $V(q)=m\omega_0^2q^2/2$, i.e. Harmonic oscillator. A saddle point approximation of such an action including gaussian fluctuations around the classical path turns out to be exact. You can then expand the fluctuations in as $\delta q=\sum_n c_n \sin(\omega_n x)$ where $\omega_n=\pi n/(t_f-t_i)$. Eventually you get something like this

$$\int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}\propto e^{S[q_c]}\frac{1}{\sqrt{\textrm{det}[-(\partial_t^2+\omega_0^2)]}}=e^{iS[q_c]}\frac{1}{\sqrt{\Pi_{n=1}^{\infty} (\omega_n^2-\omega_0^2)}}$$

Notice that since the infinite product is divergent this path integral goes to zero. I guess what eventually gives you a nonzero answer is that the normalization factor $N$ becomes infinite, but there is no way to compare the two divergencies. This is why its easier to evaluate the quotient of two path integrals such that the normalization factor N cancels and you will need to evaluate a quotient of determinants which is not divergent:

$$\frac{\textrm{det}[-(\partial_t^2+\omega_0^2)]}{\textrm{det}[-\partial_t^2]}=\Pi_n\frac{\omega_n^2-\omega_0^2}{\omega_n^2}$$

10. Aug 14, 2009

### maverick280857

Thanks jensa.

By the way, I worked out the math for myself, and I think

$$N = \left(\frac{m}{ih\tau}\right)^{n(n+1)/2}$$

$$N = \left(\frac{m}{ih\tau}\right)^{(n+1)/2}$$

This is because

$$\int \Pi_{j=0}^{n}\frac{dp_j}{h}\exp{\left(\frac{i\tau}{\hbar}\sum_{j=0}^{n}\left[\frac{p_{j}(q_{j+1}-q_{j})}{\tau}-\frac{p_{j}^2}{2m}-V(q_{j,avg})\right]\right)} = \left(\frac{m}{ih\tau}\right)^{(n+1)/2}\exp\left\{\frac{i\tau}{\hbar}\sum_{j=0}^{n}\left[\frac{m}{2}\left(\frac{q_{j+1}-q_{j}}{2}\right)^2 - V\right]\right\}$$

So the second integral $\int \Pi_{j=0}^{n}dq_{j}\ldots$ should give rise to the extra factor of n in the exponent.

Can someone please confirm whether my math is correct, because this would mean that there is a misprint in the book.