I am currently reading Quantum Mechanics and Path Integrals by Feynman and Hibbs. Working on problem 3.1 made me wonder why the 1D free particle kernel: $$ K_0(b,a) = \sqrt\frac{m}{2\pi i \hbar(t_a - t_b)} \exp \left(\frac{im(x_b - x_a)^2}{2\hbar (t_b - t_a)} \right)$$ has dimensions $1/\text{length}$ .

More generally this kernel has dimensions $1/(\text{length})^d$ , where $d$ is the number of dimensions of the system. Why is that?

On the other hand it is stated, that the kernel can be interpreted as a probability amplitude. In my understanding this would imply its dimensions to be $\sqrt{1/(\text{length})^d}$ in position space, because the kernels absolute square can be interpreted as a probability density.

Is there a physical interpretation for the normalizing factor $\sqrt{m/2\pi i \hbar(t_a - t_b)} $ other than "fixing" the equivalence to the Schrödinger equation?

To clarify my question, how can a probability amplitude have dimensions $1/(\text{length})^d$ ? Its absolute square would have dimensions $1/(\text{length})^{2d}$ which are not the proper dimensions for a probability density in my understanding.


This question has many parts, but I will try to address them all. One question you ask is why the kernel has dimensions of $\text{length}^{-d}$. Another question you asked is how the kernel can be interpreted as a probability amplitude. The third thing you asked for is a physical interpretation of the prefactor in the kernel. I will start with a review of where the kernel comes from

Review of kernel

As a toy example let's consider a piece of metal that has a temperature profile $\newcommand{\bx}{\mathbf{x}}T_0(\bx)$ as a function of position $\bx$ at time zero. The temperature at later times $t$ is described by the equation $$\partial_t T(\bx,t) = k\partial_{xx} T(\bx,t).$$

Given this equation, the temperature at a point $\newcommand{\by}{\mathbf{y}}\by$ at the time $t$ is given by $$T(\by,t)=\int \frac{1}{(4 \pi k t)^{d/2}} \exp \left(\frac{|\bx -\by|^2}{4kt}\right)T_0(\bx)d\bx\equiv\int K(\by,\bx;t)T(\bx)d\bx.$$

We have defined the kernel $K(\by,\bx;t)$ to be the function $\frac{1}{(4 \pi k t)^{d/2}} \exp \left(\frac{|\bx -\by|^2}{4kt}\right)$. Notice that since $k$ has dimensions of $\text{length}^2/\text{time}$, the function $K$ has dimensions of $\text{length}^{-d}$. It is good that the kernel these dimensions, because looking at the equation $T(\by,t)=\int K(\by,\bx;t)T(\bx)d\bx,$ if the $T$ on the right hand side is to have the same units as the $T$ from the left hand side, the dimensions of $K$ out to cancel the $d$ powers of length that result from doing the integral. That is, $K$ must have units of $\text{length}^{-d}$. In general, regardless what the dimensions of $T$ are, if the kernel $K$ is being integrated over space, then it must have dimensions of $\text{length}^{-d}$. Now it should be obvious why the kernel in your case has dimensions of $\text{length}^{-d}$.

Why the kernel has dimensions of $\text{length}^{-d}$

If $\psi_0(\bx)$ is a probability amplitude at $t=0$, having units of $\text{length}^{-d/2}$, and $\psi_t(\by)$ is a probability amplitude at time $t$, also having units of $\text{length}^{-d/2}$, and if the two are related by $\psi_t(\by)=\int K(\by,\bx;t)\psi_0(\bx)d\bx,$ then $K$ must have units of $\text{length}^{-d}$ in order for the dimensions to work out.

Interpreting the kernel itself as a solution

Next, I will address how the kernel can itself can be viewed as a solution. This seems counter-intuitive at this point because wavefunctions must have dimensions of $\text{length}^{-d/2}$, but we saw that the kernel has dimensions of $\text{length}^{-d}$. Let's go back to the example of the heat equation. Let's consider an initial temperature profile $T_0(\bx)$ that is only nonzero in a small region of volume $V$. If the volume $V$ is made smaller and smaller, then temperature will be zero everywhere at later times because the influence of the smaller region sill go to zero. Unless, that is, if the temperature within that region gets higher and higher as the region gets smaller. The limiting function, a function that is nonzero only on an infinitely small region, but that has an infinite value at that small region, is called a delta function. If we choose our initial temperature $T_0(\bx)$ to be one of these delta functions, with the inifinitely small region centered at $\mathbf{0}$, then the temperature $T(\by,t)$ at a time $t$ later is given by $\int K(\by,\bx;t)\delta(\bx)d\bx$, which is equal to $K(\by,\mathbf{0};t)$.

There is a problem with what I said above. We used a delta function as our $T_0$, but a delta function has dimenions of inverse volume, while our initial temperature profile should have dimensions of temperature. To get a true initial temperature profile, we should multiple the delta function by a constant with appropriate units (volume times temperature). The solution for the temperature at a later time would then be not just $K$, but $K$ times this same constant. Multiplying by the dimensionful constant changes the units, but does not change the spatial profile of the solution, so while $K$ doesn't have the right units to be the solution, it does have the same spatial profile of the solution for a sharply concentrated initial temperature profile.

Let's see how this works with quantum mechanics. In this case, a delta function represents the wavefunction of a particle with definite position. However, a delta function has units of inverse volume, while a wavefunction should have dimensions of square root of inverse volume. In analogy with the temperature example, the delta function should be multiplied by a constant having dimensions of square root of volume to give the appropriate dimensions for a wavefunction. Accordingly, the kernel should be multiplied by a constant of units of square root of volume. Since the kernel has units of inverse volume, this multiplicaiton gives the appropriate units of inverse square root of volume. However, the spatial profile of the wavefunction is the same whether or not you include this constant. I think this explains your second question

Interpretation of prefactor

The third thing you asked about is the meaning of the prefactor, which goes as $1/\sqrt{t}$. Since the initial state is a delta function, the momentum is completely undetermined. That is, you can think of the initial state as a superposition of all momenta. As the state evolves from the initial value, you can thihk of the wavefunction being a quantum superposition of the particle moving away from the origin at every constant velocity. So the wavefunction represents a unifrom expansion. If you have a uniform expansion of a fixed amount of mass, you expect the density to be inversely proportional to time. Since the density is inversely propotional to time, the wavefunction must be invsersely proportional to the square root of time.

  • $\begingroup$ Thank you for taking the time to write such a thorough answer, it really helped my understanding. So in conclusion $K(x,y;T) = \text{<y,T|x,0>}$ is a bit sloppy and should read $K(x,y;T) \propto \text{<y,T|x,0>}$ ? (the same goes for $P(b,a) = |K(b,a)|^2$ as stated in the book) Can the dimensionful constant you mentioned be interpreted as the uncertanty of the initial position $x_a$? (Or does it have no meaning?) $\endgroup$ – Zandorath Jan 4 '18 at 12:34
  • $\begingroup$ @Zandorath If you have a wavefunction that is constant in a volume $V_0$, and zero everywhere else, the value of the constant must be $V_0^{-1/2}$ to guarantee proper normalization. However, a delta function is the limit of a function having the value $V_0^{-1}$ as $V_0 \to 0$. Therefore the wavefunction of a localized particle is $V_0^{1/2}$ times a delta function. So you can think of the prefactor as being the square root of the volume to which the particle is confined. (No particle will truly be a pure delta function after all.) $\endgroup$ – Brian Moths Jan 4 '18 at 13:29
  • $\begingroup$ When you write: "So you can think of the prefactor as being the square root of the volume to which the particle is confined." You mean the dimensionful constant from your original post and the volume to which the particle is initially confined, right? $\endgroup$ – Zandorath Jan 4 '18 at 14:00
  • $\begingroup$ @Zandorath Yes, I mean the dimensionful constant you would need to multiply the delta function by. $\endgroup$ – Brian Moths Jan 4 '18 at 19:19

This is a consequence of $K(a,b)=\langle x_a,t_a|x_b,t_b\rangle$, which you quote, and the requirement $\displaystyle \int\!\text{d}x\,|x,t\rangle\langle x,t|=1$ (which generalizes to $\displaystyle \int\!\text{d}^dx\,|\vec x,t\rangle\langle\vec x,t|=1$ in $d$ spatial dimensions). In principle, you could label the states by some dimensionless "scale factor", but this is not the path the majority of physicists follows.

  • $\begingroup$ I guess I am kind of split between two definitions of the kernel. Firstly, the kernel gives the propability amplitude by adding up the contributions of all paths (implies $\sqrt\frac{1}{\text{length}}$). Secondly, it describes the time evolution of a wave function, when used as an integral kernel (implies $\frac{1}{\text{length}}$) $\endgroup$ – Zandorath Jan 3 '18 at 12:35

The main point is (as Ref. 1 mentions in Problem 3.1) that the probability distribution (coming from the path integral) is only relative, i.e. its normalization is unphysical over an unbounded position space $\mathbb{R}^d$. See also this and this related Phys.SE posts and links therein.


  1. R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965, Problem 3.1.
  • $\begingroup$ Thanks for your answer, it really helped me. When you say the normalization $\sqrt\frac{m}{2\pi i \hbar(t_a - t_b)}$ is unphysical, is that the answer to the question: What does the particular normalization mean? (Problem 3.1). Moreover is this always the case? $\endgroup$ – Zandorath Jan 3 '18 at 23:38

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