# I What is the physical meaning of <p>?

#### blackbeans

Summary
This is my first semester of quantum mechanics, reading Griffith's intro book, and I'm confused. I seek to clarify some misconceptions/lack of conceptions regarding <x> and <p>. I can't bring myself to move forward if I can't get a clear physical description of these two. How am I going to understand energy if I can't understand momentum. Losing sleep. A lot to unpack here, so I really appreciate the help anyone can spare.
Here's my understanding. Please let me know what I'm missing or what I've misunderstood. <x> represents the average of position measurements on some N, identically prepared particles. If those measurements are thought of as a distribution, then that distribution's form should match the wave function's probability density; the bigger N is, the better the match. If this is true, then any laboratory measurement of <x> (as described above) should be consistent with the <x> that is mathematically calculated. To me, <x> is a sort of composite image of a particle's position at some instant in time. I don't understand why <x> is an important quantity? It's astonishing that It can be measured and i'm sure it forms a foundation of more interesting theory I'll learn later. But, is that the limit of <x>'s function? Or is there something more insightful about <x> that I can understand now? What does <x> enable us to do/understand? Also, how is a particle's position measured in experiment, and is it precise?

<x> can change with respect to time, so the derivative of <x> can yield some nonzero value. By definition d<x>/dt which I'll call <v> represents the instantaneous rate of change of <x>. It's a measure of how fast the average of x changes, and seems unrelated to classical velocity. Griffiths makes it sound that particle's don't really have velocities anyway. That they can't be measured. It seems that <v> is not like <x> in that <x> can be verified by experiment.

What really, really confuses me is <p> = d<x>/dt * m. If my definition of <v> is correct, then it seems reasonable to want to measure it. But <p> seems like a weird thing to calculate. Perhaps it's because I'm thinking classically, but it's difficult not to when the operators look so similar to their classical counterparts. Is <p> what we would expect if we measured the momentum in N identically prepared experiments and took their average (how is this possible if <v> can't be measured in that way)? Is it a measure of the rate of change of <x> scaled by mass? Or are those two things one and the same?

Thanks a bunch.

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#### PeterDonis

Mentor
Please let me know what I'm missing or what I've misunderstood.
In your post you left out some key steps: before even trying to understand what <x> and <p> mean, you need to first understand what <> means and what $x$ and $p$ (without the <> around them) mean.

First, $x$ and $p$ should really be written with hats on them: $\hat{x}$ and $\hat{p}$. The hat tells you that these are operators. What is an operator? It's an operation that you apply to a state to get another state. Physically, operators, or more precisely self-adjoint operators, correspond to measurements you can make on a quantum system, which we'll consider here to be a single free particle. (We'll pass over various technicalities involved in the definition of "self-adjoint" and the fact that the operators $\hat{x}$ and $\hat{p}$ can't be normalized.) The operator $\hat{x}$ corresponds to measuring the particle's position. The operator $\hat{p}$ corresponds to measuring the particle's momentum. Note that these operators say nothing specific about how you make the measurements.

Next, the <> around an operator means taking its expectation value. More precisely, it means taking its expectation value with respect to some particular state. Physically, this can be thought of more or less as you described <x>: if you have a large number of particles all prepared in the same state, and you make a position measurement on each one of them independently, the average of all the measurements should be close to <x> calculated for the state the particles were prepared in. In mathematical notation, if all the particles were prepared in the same state $| \psi \rangle$, then the average of all the position measurements is predicted to be $\langle \hat{x} \rangle = \langle \psi | \hat{x} | \psi \rangle$.

You can do the same thing with any operator, for example the operator $\hat{p}$, which, as noted above, corresponds to measuring the particle's momentum. So, again, if you have a large number of particles all prepared in the same state $| \psi \rangle$, and you measure the momentum of each one of them independently, the average of all the momentum measurements is predicted to be $\langle \hat{p} \rangle = \langle \psi | \hat{p} | \psi \rangle$.

The equation

$$\langle \hat{p} \rangle = m \frac{d}{dt} \langle \hat{x} \rangle$$

is a special case of what is known as Ehrenfest's Theorem. All it is telling you is that, if you do both of the experiments I described above, making sure that the prepared state of the particles $| \psi \rangle$ is the same in both, and keep repeating this over time, the averages you calculate will be related as the equation says. You have to keep repeating the experiments over time in order to calculate the time derivative of $\langle \hat{x} \rangle$.

By definition d<x>/dt which I'll call <v> represents the instantaneous rate of change of <x>.
You can indeed make this definition, but, as you seem to have realized, this does not mean that there must be some operator $\hat{v}$ that $\langle v \rangle$ is the expectation value of. In this particular case, it turns out that there is: given Ehrenfest's Theorem, you can write $\langle v \rangle = \langle \hat{p} \rangle / m$. And since $\hat{p}$ itself is a known operator (not the time derivative of one), you can define a "velocity operator" $\hat{v} = \hat{p} / m$. Physically, this would correspond to measuring velocity by measuring momentum and then dividing the result by the particle's mass. Then you could write

$$\langle \hat{v} \rangle = \frac{d}{dt} \langle \hat{x} \rangle$$

which is just another way of stating Ehrenfest's Theorem for this case. However, this trick does not work in all cases, which is why you usually see the momentum operator $\hat{p}$ in QM textbooks.

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#### Prathyush

If you want to develop your intuition you should also read Feynman and Hibbs.

#### vanhees71

Gold Member
Or, even better, vol. III of the Feynman Lectures on Physics. I don't think, it's good if you try to learn all the different formalisms in which QT can be formulated at once. The path-integral formalism is very nice to do some calculations, but imho it's not a good method for introductory QT, because it's very definition is so complicated, and the evaluation of even some quite straight-forward things is very cumbersome.

I think for non-relativistic QM there are two approaches, which are most straight forward to get started:

(a) The traditional wave-mechanics approach. There you use the Einstein-de Broglie idea to describe what's historically called "wave-particle dualism" and use it as a heuristic way to derive the corresponding wave equation. Famously, when Schrödinger gave a seminar at the ETH on de Broglie's thesis (very well evaluated by Einstein, who was asked by Langevin, how to judge the thesis ;-)), Debye (being an off-spring of the no-nonsense mathematical approach of the Sommerfeld school) made the remark that if there are waves, there should be a wave equation. Schrödinger obviously took this at heart, went on summer vacation with one of his muses (it's not sure who it was) and created quantum mechanics as an eigenvalue problem. In addition you only need to learn the meaning of the wave function as a "probablitiy amplitude", which is due to a footnote by Born in his paper on scattering theory.

(b) Dirac's respresentation-free operator approach. There you take "canonical quantization" as a heuristics, i.e., you make the observables known from classical mechanics (phase-space functions) operators and use as a correspondence rule "Poisson brackets $\leftrightarrow$ commutators times $-\mathrm{i}$). The disadvantage of this is that it's pretty formal to begin with. The advantage is you can start with the most simple conceivable examples of two-level systems (polarization states of photons or spin states of a spin-1/2 particle like the electron).

The path-integral formalism can then be derived after the operator formalism is settled, first by going to a "time lattice" and taking the continuum limit (which is quite cumbersome) and then using some advanced functional techniques like the Schwinger-time formalism or the heat-kernel formalism, for which however you need to be pretty firm with function theory, contour integration, and analytical continuation.

I'm a bit uncertain, whether Griffiths is really a good introductory textbook on QM, because whenever a student posts in this forum, the standard introductory sentence is exactly the one of this thread: "This is my first semester of quantum mechanics, reading Griffith's intro book, and I'm confused." Though some confusion is normal learning QM, it seems that there's more confusion than necessary when Griffiths's book is used. My favorite as an intro book, inclined more to possibility (b),

J. J. Sakurai, S. Tuan, Modern Quantum Mechanics, Addison Wesley (1993).

For the wave-function approach a masterpiece still is

W. Pauli, General Principles of Quantum Mechanics, Springer-Verlag, Berlin, Heidelberg (1980).

That said, let me try to answer the questions concerning expectation values. If you use the wave-mechanics approach (I think Griffiths chooses it for his book), then you should have learnt that a (pure) quantum state of a single particle, neglecting spin, is represented by a wave function $\psi(t,\vec{x})$ (modulo an unimportant phase factor). These "wave functions" build a Hilbert space, i.e., with any two wave functions also an arbitrary linear combination is again a wave function, and there's defined a scalar product on this vector space of functions, given by
$$\langle \psi(t)|\phi(t) \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x \psi^*(t,\vec{x}) \phi(t,\vec{x}).$$
The observables are represented by self-adjoint operators, where self-adjoint means that for any two wave functions (in an appropriate dense subspace of the Hilbert space, where the operators are well defined)
$$\langle \psi(t)|\hat{O} \phi(t) \rangle=\langle \hat{O} \psi(t)|\phi(t) \rangle,$$
which means written in terms of integrals
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x \psi^*(t,\vec{x}) \hat{O} \phi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x [\hat{O} \psi(t,\vec{x})]^* \phi(t,\vec{x}).$$
Now, using the Einstein-de Broglie argument for the correspondence between "wave properties" and "particle properties", using plain waves (which are strictly speaking not wave functions in the above sense since they are not square integrable) you come to the conclusion that position and momentum are represented by the self-adjoint operators
$$\hat{\vec{x}} \psi(t,\vec{x})=\vec{x} \psi(t,\vec{x}), \quad \hat{\vec{p}} \psi(t,\vec{x}) = \frac{\hbar}{\mathrm{i}} \vec{\nabla} \psi(t,\vec{x}).$$
The meaning of the wave function is that
$$P(t,\vec{x}|\psi)=|\psi(t,\vec{x})|^2$$
is the probability distribution for finding a particle at position $\vec{x}$ when measured at time $t$, provided the particle is prepared in a state described by the wave function $\psi(t,\vec{x})$.

Now it's already easy to see that the expecation value for position must be
$$\langle \vec{x} \rangle(t)=\int_{\mathbb{R}} \mathrm{d}^3 x x P(t,\vec{x}|\psi) = \int_{\mathbb{R}} \mathrm{d}^3 x \psi^*(t,\vec{x}) \hat{x} \psi(t,\vec{x}),$$
where we assume that the wave function is properly normalized to 1:
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x |\psi(t,\vec{x})|^2=1.$$
This suggests that for the momentum you have
$$\langle \vec{p} \rangle(t)=\int_{\mathbb{R}} \mathrm{d}^3 x \psi^*(t,\vec{x}) \hat{\vec{p}} \psi(t,\vec{x}) = \int_{\mathbb{R}} \mathrm{d}^3 x \psi^*(t,\vec{x}) \frac{\hbar}{\mathrm{i}} \vec{\nabla} \psi(t, \vec{x}).$$
Finally, from the Einstein-de Broglie relation between energy and momentum, you get the Schrödinger wave equation,
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),$$
where $\hat{H}$ is the operator for the Hamilton function (or the energy) of the particle. E.g. for a particle moving in an external field with a potential you have
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2 + V(\hat{\vec{x}}).$$
Plugging this into the integral for the expectation values, you indeed get "Ehrenfest's theorem", i.e.,
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle x \rangle(t) = \frac{1}{m} \langle p \rangle(t), \quad \frac{\mathrm{d}}{\mathrm{d} t} \langle p \rangle(t)=-\langle \vec{\nabla} V(\hat{\vec{x}}) \rangle(t).$$
This makes the above conjecture concerning the momentum operator plausible.

What also follows more generally for any observable described by a self-adjoint operator $\hat{O}=\hat{O}(\hat{x},\hat{p})$ by plugging it into the equation for the expectation value is that the operator describing the time derivative of this observable is
$$\mathring{\hat{O}}(\hat{\vec{x}},\hat{\vec{p}})=\frac{1}{\mathrm{i} \hbar} [\hat{O}(\hat{\vec{x}},\hat{\vec{p}}),\hat{H}],$$
and this indeed resembles the corresponding equation of the Hamiltonian formalism in classical mechanics, when expressed in terms of Poisson brackets.

One can indeed easily show that
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle O(\hat{\vec{x}},\hat{\vec{p}}) \rangle(t) = \langle \mathring{\hat{O}}(\hat{\vec{x}},\hat{\vec{p}}) \rangle.$$

"What is the physical meaning of <p>?"

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