B Question about squares of operators

[etotheipi]

The magnitude of the momentum $p$ satisfies $p^2 = p_x^2 + p_y^2 + p_z^2$ and this implies the operator equation $\hat{p}^2 = \hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2$, so we can say that $\hat{p}^2 = -\hbar^2 (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})$.

I had two questions. Firstly, is $\hat{p}^2$ equivalent to $\hat{p^2}$, or does the latter not make any sense?

And secondly, we can say $\hat{p}^2 \psi = p^2 \psi$, but does it make any sense to talk about the 'unsquared' version of the operator for the magnitude of the momentum, $\hat{p}$? With the squared $\hat{p}^2$ the $\psi$ distributes into the bracket, but I'm not sure if it is valid to write $\hat{p} \psi = p \psi$.

Thank you, and sorry if these are silly questions !

 

[jfizzix]

It is possible to construct an operator that is the square root of \hat{p}^{2} and call it \hat{p}, and such an operator would satisfy the the relation I think you're stating.

If I read your second equation:
\hat{p}^{2}\psi=p^{2}\psi
correctly, this is saying that applying the operator \hat{p}^{2} to \psi is equaivalent to multiplying the number p^{2} to \psi. This is not generally true, and will only be true for special instances of \psi where \psi is an eigenstate of \hat{p}^{2}.

In general, \psi will be in a superposition of multiple eigenstates of \hat{p}^{2}, so that when measuring \hat{p}^{2} of the state \psi, multiple possible values of p^{2} are possible.

 

[PeroK]

The magnitude of the momentum $p$ satisfies $p^2 = p_x^2 + p_y^2 + p_z^2$ and this implies the operator equation $\hat{p}^2 = \hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2$, so we can say that $\hat{p}^2 = -\hbar^2 (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})$.

I had two questions. Firstly, is $\hat{p}^2$ equivalent to $\hat{p^2}$, or does the latter not make any sense?

And secondly, we can say $\hat{p}^2 \psi = p^2 \psi$, but does it make any sense to talk about the 'unsquared' version of the operator for the magnitude of the momentum, $\hat{p}$? With the squared $\hat{p}^2$ the $\psi$ distributes into the bracket, but I'm not sure if it is valid to write $\hat{p} \psi = p \psi$.

Thank you, and sorry if these are silly questions !

In general, if you have observables $\mathbf{Q_1, Q_2}$, repesented by operators $\mathbf{\hat Q_1, \hat Q_2}$, then the observable $\mathbf{Q_1 \cdot Q_2}$ is represented by the operator $\mathbf{\hat Q_1 \cdot \hat Q_2}$.

In general, it's better to write $\mathbf{\hat p^2}$, as this is explicity $\mathbf{ \hat p \cdot \hat p}$.

 

[etotheipi]

Thanks for your replies! First off, I'm guessing then that $\mathbf{\hat{p^2}}$ is meaningless, so I will forget that I ever saw it from now on!

This is not generally true, and will only be true for special instances of \psi where \psi is an eigenstate of \hat{p}^{2}.

Yes, sorry, I should have written that part explicitly! The assumption was that they were eigenvalues

In general, if you have observables $\mathbf{Q_1, Q_2}$, repesented by operators $\mathbf{\hat Q_1, \hat Q_2}$, then the observable $\mathbf{Q_1 \cdot Q_2}$ is represented by the operator $\mathbf{\hat Q_1 \cdot \hat Q_2}$.

In general, it's better to write $\mathbf{\hat p^2}$, as this is explicity $\mathbf{ \hat p \cdot \hat p}$.

On another note, can you always transform an equation into an operator equation? I.e. if we have $T = \frac{\mathbf{p}\cdot \mathbf{p}}{2m}$, does it always follow that $\hat{T} = \frac{\hat{\mathbf{p}} \cdot \hat{\mathbf{p}}}{2m}$?

From what I could tell from playing around with a few examples, I concluded that it seemed reasonable, but I don't really know if it generalises!

 

[PeroK]

On another note, can you always transform an equation into an operator equation? I.e. if we have $T = \frac{\mathbf{p}\cdot \mathbf{p}}{2m}$, does it always follow that $\hat{T} = \frac{\hat{\mathbf{p}} \cdot \hat{\mathbf{p}}}{2m}$?

From what I could tell from playing around with a few examples, I concluded that it seemed reasonable, but I don't really know if it generalises!

That's a general rule that applies to any function of the observables. If you have a function $f(x, p)$, say, then the associated operator is $f(\hat x, \hat p)$.

Note that things really ought to go the other way. I.e. we ought to work things out quantum mechanically and then show that the macroscopic operator is related to the quantum operator in this way. But, we have no direct experience of quantum mechanics to guide us, so generally we take what we know from CM and work backwards!

In other words, you can't really prove that $\mathbf p \rightarrow -i\hbar \nabla$. But, if you could work QM out first, then you could show that $-i\hbar \nabla \rightarrow \mathbf p$

 

[etotheipi]

In other words, you can't really prove that $\mathbf p \rightarrow -i\hbar \nabla$. But, if you could work QM out first, then you could show that $-i\hbar \nabla \rightarrow \mathbf p$

I hadn't realized this! The way I saw was to say that $\psi = Ae^{\frac{i}{\hbar}(px-Et)}$, then $\frac{\partial}{\partial x}$ both sides and rearrange to get the operator. But now I think about it, we had to have assumed the solution first.

That's a general rule that applies to any function of the observables. If you have a function $f(x, p)$, say, then the associated operator is $f(\hat x, \hat p)$.

That's reassuring; thanks for clarifying!

 

[PeroK]

I hadn't realized this! The way I saw was to say that $\psi = Ae^{\frac{i}{\hbar}(px-Et)}$, then $\frac{\partial}{\partial x}$ both sides and rearrange to get the operator. But now I think about it, we had to have assumed the solution first.
That's reassuring; thanks for clarifying!

If you want a coherent introduction to QM, you can't do much better than:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

 

[etotheipi]

If you want a coherent introduction to QM, you can't do much better than:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

Thanks, I'll start looking through those! At the moment it's a bit tricky to decide what the best route into the subject is since all the notes I've seen start out at different places and focus on different topics.

 

[etotheipi]

That's a general rule that applies to any function of the observables. If you have a function $f(x, p)$, say, then the associated operator is $f(\hat x, \hat p)$.

Just found it in the notes you linked, apparently the rule is called "canonical quantisation". Sounds fancy!

 

[vanhees71]

In other words, you can't really prove that $\mathbf p \rightarrow -i\hbar \nabla$. But, if you could work QM out first, then you could show that $-i\hbar \nabla \rightarrow \mathbf p$

This you cannot say in this general way. It depends on, where you start, whether something is provable or not.

I think the best way to learn quantum mechanics is to first learn classical mechanics in the analytical formulation, i.e., the Hamilton principle of least action in the Hamilton way and then a good deal of elementary Lie-group and Lie-algebra theory in terms of Poisson brackets. The key point is to understand Noether's theorem.

Then, what you indeed cannot derive starting from classical mechanics is of course quantum mechanics, because classical mechanics is an approximation of quantum mechanics for macroscopic observables on many-body systems. So you need some heuristics to understand, why there's a Hilbert space of states and operators that represent observables.

If you are that far, it's quite clear, how to get educated guesses for the operators representing a given observable known from classical mechanics: The 10 conservation laws of Newtonian mechanics connect the observables with Lie symmetries, and that's why the operators build a unitary representation of the corresponding symmetries on Hilbert space.

Now momentum is the conserved quantity related to spatial translations, i.e., it describes "infinitesimal spatial translations". Now take the representation of QM in terms of wave mechanics and check, how spatial translations are represented in terms of the wave equation. The most simple idea is to make the wave function a scalar field under translations, i.e., if the translation is described by $\vec{x}'=\vec{x}-\vec{a}$, then
$$\psi'(\vec{x}')=\psi(\vec{x})=\psi(\vec{x}'+\vec{a}).$$
Now make this translation infinitesimal, so that
$$\psi'(\vec{x}')=\psi(\vec{x}'+\delta \vec{a})=\psi(\vec{x}')+\delta \vec{a} \cdot \vec{\nabla} \psi(\vec{x}')+\mathcal{O}(\delta \vec{a}^2).$$
Now $\hat{\vec{p}}$ should precisely do such in infinitesimal translation, i.e.,
$$\mathrm{i} \delta \vec{a} \hat{\vec{p}} \psi(\vec{x}) \stackrel{!}{=} \delta \vec{a} \cdot \vec{\nabla} \psi(\vec{x}).$$
Comparing both sides gives
$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}.$$
Note that I used natural units with $\hbar=1$ here. It's easy to implement the $\hbar$ from dimensional reasoning:
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}.$$
That I introduced the imaginary $\mathrm{i}$ is for the reason that we want the generators of the symmetry and thus the operator representing momentum to be self-adjoint rather than anti-self-adjoint, because that's more convenient in the formalism.

Now from the infinitesimal symmetry generators you get the finite transformations by the operator exponential function, i.e., the unitary translation operator by definition fulfills
$$\vec{\nabla}_a \hat{U}(\vec{a})=\mathrm{i} \hat{\vec{p}} \hat{U}(\vec{a}),$$
and with the initial condition $\hat{U}(0)=\hat{1}$ this gives
$$\hat{U}(\vec{a})=\exp(\mathrm{i} \vec{a} \cdot \hat{\vec{p}}) .$$
Now indeed you get with our above result $\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}$
$$\hat{U}(\vec{a}) \psi(\vec{x})=\exp(\vec{a} \cdot \vec{\nabla} \psi(\vec{x}) = \sum_{k=0}^{\infty} \frac{1}{k!} (\vec{a} \cdot \vec{\nabla}) \psi(\vec{x})=\psi(\vec{x}+\vec{a}),$$
where in the last step I have used the Taylor expansion of functions.

As you see, indeed $\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}$ is the generator of translations in quantum theory in the same way it is in classical mechanics in Hamiltonian formulation.

This heuristic connection brought Dirac to his formulation of quantum theory, by translating the classical quantitities and Poisson brackets to quantum descriptions by making the classical quantities self-adjoint operators on a Hilbert space and the Poisson brackets commutators (up to a factor $\mathrm{i}$). Then the starting point to define position and momentum operators are the canonical Poisson brackets translating into the canonical commutation relations for the operators in quantum theory,
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hat{1}.$$
You can also use this "Heisenberg algebra" to derive the above result for the momentum operator using the position representation, i.e., the description of the state vectors in terms of "components" wrt. the position eigenbasis: $\psi(\vec{x})=\langle \vec{x}|\psi \rangle$.

 

[PeroK]

This you cannot say in this general way. It depends on, where you start, whether something is provable or not.

My point was more as follows. If we study the mechanics of a snooker ball, we cannot say what the snooker ball is made of; and, assuming it has elementary particles, how they behave. Even as late as 1900 it was not universally accepted that there were atoms and that matter was not a continuous distribution of something. Classical mechanics cannot fully determine the mechanics of elementary particles.

But, if we know how the elementary particles behave, then we can (at least in principle) explain how macroscopic objects behave. In principle, QM should fully determine classical mechanics.

 

[Truecrimson]

That's a general rule that applies to any function of the observables. If you have a function $f(x, p)$, say, then the associated operator is $f(\hat x, \hat p)$.

Would there be a problem with products of non-commuting operators? Classically, $f(x,p) = x^2p^2 = q xp^2 x + (1-q) px^2p$ for any $0\le q \le 1$ but different choices of $q$ lead to different observables.

 

[Demystifier]

Would there be a problem with products of non-commuting operators? Classically, $f(x,p) = x^2p^2 = q xp^2 x + (1-q) px^2p$ for any $0\le q \le 1$ but different choices of $q$ lead to different observables.

Yes, this is the well known problem of ordering ambiguity in QM. It is related to an even better known notion of vacuum energy and normal ordering. Ultimately, this is also an origin of UV divergences and a need for renormalization in continuum QFT.

 

[vanhees71]

Yes, and if in doubt, how to translate a "classical expression" to the "quantum expression", one has to think about the physical meaning of the observable and hopefully find some group-theoretical argument based on symmetries, what's the right expression to use in the quantum theory. There's no mathematically unique mapping from classical to quantum mechanics, because classical mechanics is only an approximation to quantum mechanics.

The UV divergences in QFT are still another issue. There the problem is how to deal with distribution valued field operators and local products of them.

 

[PGaccount]

The issue of the relationship of classical mechanics to quantum mechanics is a peculiar one. The quantum mechanical amplitudes are defined by φ ~ exp iS, where S is defined by

S = ∫dt L

L = 1/2 mv² - U(x)

This L is the difference of the classical kinetic and potential energies. We must first have the expression for L which defines the classical theory, and only then do we get the amplitudes.dream Another way to see why classical mechanics is necessary for the definition of quantum mechanics is the fact that we must necessarily give a classical account of the measuring bodies.

 

[DrDu]

I had two questions. Firstly, is $\hat{p}^2$ equivalent to $\hat{p^2}$, or does the latter not make any sense?

It is the other way round, the first expression makes no sense! $\hat{p}$ is an unbounded operator, which means that it is not defined on all elements of the Hilbert space. Specifically it is not defined on all of the range of $\hat{p}$. E.g. $\hat{p} \exp(-k|x|)$ is not in the domain of $\hat{p}$. Hence, $\hat{p}^2$ has a smaller domain than $\hat{p}$ and it turns out that the remaining domain is not sufficient to make it a self-adjoint operator.

 

[vanhees71]

The function $\exp(\mathrm{i} k |x|)$ is in no operator's domain either. It's not even an element of the Hilbert space $\mathrm{L}^2(\mathbb{R})$ to begin with.

 

[DrDu]

The function $\exp(\mathrm{i} k |x|)$ is in no operator's domain either. It's not even an element of the Hilbert space $\mathrm{L}^2(\mathbb{R})$ to begin with.

Of course you are right. I wanted to write $\exp(-k|x|)$. Corrected this.