Heisenberg uncertainty principle and the canonical momentum operator

[wnvl2]

Heisenberg uncertainty principle relates to the limitations on the precision with which certain pairs of physical properties (like position and momentum) of a particle can be simultaneously known. The uncertainty relations for position and momentum as operators arise from the non-commutative nature of these operators. Specifically, the position $\hat{x}$ and momentum $\hat{p}$ operators do not commute, which leads to a fundamental limit on the precision with which these quantities can be simultaneously measured.

Can these relations in al situations be applied for the momentum operator
$$
\hat{p} = -i\hbar \frac{\partial}{\partial x},
$$
or should in some cases the canonical momentum operator be used?

For instance if I want to apply the Heisenberg uncertainty relations to the kernal of an atom is it simply with operator
$$
\hat{p} = -i\hbar \frac{\partial}{\partial x},
$$
or do I have to us a more complex operator for the canonical momentum?

 

[anuttarasammyak]

Not only free space but also in general field, that formula is applicable.
For motions of charged particles momentum is transformed as
$$p\rightarrow p-eA$$
but it does not affect that formula, I think. Corrections by experts are welcome.

 

[pines-demon]

Do you know Poisson's brackets? This can be used to make canonical quantization clear.
Edit: also there is a generalized uncertainty principle that you should check.

 

[wnvl2]

I know Poisson brackets. I will check the generalized uncertainty principle.

 

[pines-demon]

I know Poisson brackets. I will check the generalized uncertainty principle.

So basically if the Poisson brackets between your canonical variables are $\{q,p\}=1$ then you quantize by fixing that the operators must not commutate and be given by $[\hat{q},\hat{p}]=i\hbar$. So yeah, it is the canonical momentum in the position base that is written as $-i\hbar \partial/\partial q$ in the position (whatever $q$ and $p$ are). The same happens with angle $\theta$ and angular momentum around that angle, where $L_z=-i\hbar\partial/\partial\theta$.

Check Conjugate variables (Wikipedia)

 

[pines-demon]

For a less trivial example you can check the quantum LC circuit (that is sometimes used to model superconductive circuits). In that you quantize the charge in the capacitor and its conjugate momentum (voltage).

 

[fresh_42]

Check Conjugate variables (Wikipedia)

Fun fact: the German Wikipedia page titles "complementary observables".

 

[PeterDonis]

it is the canonical momentum in the position base that is written as $-i\hbar \partial/\partial q$ in the position (whatever $q$ and $p$ are).

Only for a free particle. For a particle in a vector potential $A$, the canonical momentum has an extra term $e A$, where $e$ is the charge. That means the operator $- i \hbar \partial / \partial q$ and the canonical momentum operator are no longer the same. And if $A$ does not commute with the position operator, there will be different Heisenberg uncertainty relations with position for the two operators.

 

[pines-demon]

Only for a free particle. For a particle in a vector potential $A$, the canonical momentum has an extra term $e A$, where $e$ is the charge. That means the operator $- i \hbar \partial / \partial q$ and the canonical momentum operator are no longer the same. And if $A$ does not commute with the position operator, there will be different Heisenberg uncertainty relations with position for the two operators.

Isn't it the whole point of minimal coupling saying that in the presence of $\mathbf A(\mathbf r)$ that the canonical momentum is $\mathbf p = m \mathbf v + q\mathbf A (\mathbf r)$? So you have to replace $\mathbf p_{\text{free}} \to \mathbf p - q\mathbf A(\mathbf x)$ [compare signs] and is the new $\mathbf p$ that is written as $-i\hbar \nabla$ (and not the one representing old momentum $\mathbf p_{\text{free}}=m\mathbf v$)?

 

[DrClaude]

Isn't it the whole point of minimal coupling saying that in the presence of $\mathbf A(\mathbf r)$ that the canonical momentum is $\mathbf p = m \mathbf v + q\mathbf A (\mathbf r)$? So you have to replace $\mathbf p_{\text{free}} \to \mathbf p - q\mathbf A(\mathbf x)$ [compare signs] and is the new $\mathbf p$ that is written as $-i\hbar \nabla$ (and not the one representing old momentum $\mathbf p_{\text{free}}=m\mathbf v$)?

Correct. You still have $\mathbf{p} = -i\hbar \nabla$, but $\mathbf{p}$ is no longer $m \mathbf{v}$.

 

[pines-demon]

Correct. You still have $\mathbf{p} = -i\hbar \nabla$, but $\mathbf{p}$ is no longer $m \mathbf{v}$.

And do you agree that the $\mathbf p=-i\hbar\nabla$ that you defined IS the canonical momentum? (even if $\mathbf p\neq m\mathbf v$).

 

[DrClaude]

And do you agree that the $\mathbf p=-i\hbar\nabla$ that you defined IS the canonical momentum? (even if $\mathbf p\neq m\mathbf v$).

Yes. To quote from Sakura and Napolitano, Modern Quantum Mechanics:

Quite often $\mathbf{p}$ is called canonical momentum, as distinguished from kinematical (or mechanical) momentum, denoted by $\mathbf{\Pi}$:
$$
\mathbf{\Pi} \equiv m \frac{d \mathbf{x}}{dt} = \mathbf{p} - \frac{e \mathbf{A}}{c}.
$$

 

[PeterDonis]

You still have $\mathbf{p} = -i\hbar \nabla$

Is that the case? The canonical definition is derived from the Lagrangian: $p = \delta L / \delta \dot{q}$. That doesn't seem like the same thing.

 

[pines-demon]

Is that the case? The canonical definition is derived from the Lagrangian: $p = \delta L / \delta \dot{q}$. That doesn't seem like the same thing.

The Lagrangian is
$$L=\frac12 m\dot{\mathbf q}^2+Q\dot{\mathbf q} \cdot \mathbf A$$
the canonical momentum is
$$p_i=\frac{\partial L}{\partial \dot{q}_i}= m \dot{q}_i+Q A_i $$
Writing the Hamiltonian
$$H=\sum_i p_i \dot{q}_i-L=\frac{m \dot{q}^2 }{2}=\frac{|\mathbf p - Q \mathbf A|^2}{2m}$$,
and that $\mathbf p$ is the one that becomes $-i\hbar \nabla$.

 

[PeterDonis]

that $\mathbf p$ is the one that becomes $-i\hbar \nabla$.

How does that happen? That's my question. You can't just declare by fiat that they're the same.

 

[pines-demon]

How does that happen? That's my question. You can't just declare by fiat that they're the same.

Because that's how canonical quantization works. We are working in the Hamiltonian picture with position and canonical momentum. Have you ever solved a particle in a magnetic field (Landau levels) we certainly do NOT make $\mathbf p - Q \mathbf A=-i\hbar \nabla$.

 

[PeterDonis]

a particle in a magnetic field (Landau levels) we certainly do not make $\mathbf p - Q \mathbf A=-i\hbar \nabla$.

Ah, ok, I was misunderstanding what you said.

 

[pines-demon]

Ah, ok, I was misunderstanding what you said.

I still do not know what you meant in #8.

 

[PeterDonis]

I still do not know what you meant in #8.

The Lagrangian for a particle in a vector potential includes a term $e v \cdot A$, which leads to a term $e A$ in the canonical momentum: it becomes $p = M \dot{q} + e A$. (See, for example, Ballentine, section 11.1. I used $e$ for the charge instead of $q$ to avoid confusion with the configuration variable $q$.)

The thing that might need to be spelled out more explicitly is how we get from $p = M \dot{q} + e A$ to $p = - i \hbar \partial / \partial q$. I posted post #8, and subsequent posts, to raise that question.

 

[pines-demon]

Using the canonical quantization you get that $\{x,p\}=1\to[\hat x, \hat p ]=i\hbar$. Now, the tricky part of what you are asking is how do you get from the commutator to $p=-i\hbar \partial_x$ (in $\hat x$-basis). Here is a derivation: https://physics.stackexchange.com/q...-position-basis-from-its-commutation-relation

Apparently, the rigorous name is Stone-von Neumann theorem.

 

[PeterDonis]

Using the canonical quantization you get that $\{x,p\}=1\to[\hat x, \hat p ]=i\hbar$.

Ah, I see; it relies on the Poisson bracket--commutator correspondence. Ballentine, Section 3.6, discusses limitations of that approach to quantization, which is not the approach Ballentine takes. Ballentine's (very brief) discussion of canonical momentum makes no mention of canonical quantization, nor does it appear to me to make use of $p = - i \hbar \nabla$. I'm not sure how the latter formula would be derived in the general case using Ballentine's approach.

the rigorous name is Stone-von Neumann theorem.

That's the theorem that shows that all representations of the canonical commutation relations must be unitarily equivalent. (Note that the theorem only holds for the case of finitely many degrees of freedom; for the case of infinitely many degrees of freedom, as in quantum field theory, there are unitarily inequivalent representations, which is a major issue in QFT foundations.)

 

[pines-demon]

Ah, I see; it relies on the Poisson bracket--commutator correspondence. Ballentine, Section 3.6, discusses limitations of that approach to quantization, which is not the approach Ballentine takes. Ballentine's (very brief) discussion of canonical momentum makes no mention of canonical quantization, nor does it appear to me to make use of $p = - i \hbar \nabla$. I'm not sure how the latter formula would be derived in the general case using Ballentine's approach.


That's the theorem that shows that all representations of the canonical commutation relations must be unitarily equivalent. (Note that the theorem only holds for the case of finitely many degrees of freedom; for the case of infinitely many degrees of freedom, as in quantum field theory, there are unitarily inequivalent representations, which is a major issue in QFT foundations.)

Ballentine cites J.R. Sewell 1959 for the canonical commutation, it has a good discussion on that and its many limitations, the uniqueness of the mapping of the commutator to $-i\hbar \partial_q$ fails when not all conditions are met. Sewell contents himself with saying that it works for simplicity which is a principle often used in physics. However I also wonder what are the modern alternatives if any...

 

[DrClaude]

Ballentine cites J.R. Sewell 1959 for the canonical commutation

Could you please give the full citation of that article?

 

[pines-demon]

Could you please give the full citation of that article?

Shewell, J.R. 1959, Am. J. Phys. 27, 16

Doi: https://doi.org/10.1119/1.1934740

 

[anuttarasammyak]

Yes. To quote from Sakura and Napolitano, Modern Quantum Mechanics:

Thanks for the good information. May we regard that (which ?) one of $\Pi, p$ is mechanical momentum of a charged particle and the other is mechanical mometum of the system,i.e. a charged particle and EM field ?

 

[pines-demon]

Thanks for the good information. May we regard that (which ?) one of $\Pi, p$ is mechanical momentum of a charged particle and the other is mechanical mometum of the system,i.e. a charged particle and EM field ?

What is mechanical momentum?

 

[anuttarasammyak]

What is mechanical momentum?

I find it in the DrClaude's quote of Sakurai. I would like to understand more clearly how we should deal with momentum of canonical/kinetatical(mechanical) for a particle/em field/both(the system), and which corresponds to x-differential operator of particle.

 

[DrClaude]

Doi: https://doi.org/10.1119/1.1934740

Thanks!

Thanks for the good information. May we regard that (which ?) one of $\Pi, p$ is mechanical momentum of a charged particle and the other is mechanical mometum of the system,i.e. a charged particle and EM field ?

They both concern only the particle. The vector potential $\mathbf{A}$ and scalar potential $\phi$ correspond to an external electromagnetic field (they do not take into account the field created by the charged particle).

 

[anuttarasammyak]

Thanks. Landau, classical thery of field says

sign of A is different from Sakurai and Napolitano. It is due to conventions they refer or I take wrong correspondence of P,p in Landuau and $\Pi$, p in Sakurai ?

 

[pines-demon]

Thanks. Landau, classical thery of field says

View attachment 354579

sign of A is different from Sakurai and Napolitano. It is due to conventions they refer or I take wrong correspondence of P,p in Landuau and $\Pi$, p in Sakurai ?

Again the canonical momentum is $\mathbf P=m\mathbf v +q\mathbf A=\boldsymbol \Pi+q\mathbf A$, where $\boldsymbol \Pi$ is the ordinary momentum. All the books are consistent unless the charge is taken as negative for electrons.

 

[joneall]

I'm arriving late for this discussion, but it is just in time as I am now reading about the subject in Sakurai and Napolitano. Just to get things straighter, let me quote more from them, p. 128.

"In general, the canonical momentum p is not a gauge-invariant quantity; its numerical value depends on the particular gauge used even when we are referring to the same physical situation. In contrast, the kinematic [or mechanical] momentum $\Pi$, or $mdx/dt$ that traces the trajectory of the particle [!] is a gauge-invariant quantity... Because $p$ and $mdx/dt$ are related..., p must change to compensate for the change in A..."

Here, the relation he is referring to is $\Pi = m\frac{dx}{dt} = p - \frac{eA}{c}$.

Two pages later, they say:

"To summarize, when vector potentials in different gauges are used for the same physical situation,the corresponding state kets (or wave functions) must necessarily be different. However, only a simple change is needed; we can go from a gauge specified by $A$ to another specified by $A + \nabla\Lambda$ by merely multiplying the old ket (the old wave function) by $exp[ie\Lambda(x)/\hbar c]exp[ie\Lambda(x^{\prime})/\hbar c]$. The canonical momentum, defined as the generator of translation, is manifestly gauge dependent in the sense that its expectation value depends on the particular gauge chosen, while the kinematic [or mechanical] momentum and the probability flux are gauge invariant."

Let's see if I understand this correctly. The case considered is that of a charged particle in a magnetic field (ignoring eventual magnetic effects due to the charge).

-- That the mechanical momentum traces the trajectory is just because it is proportional to the acceleration.

-- The canonical momentum is the generator of translation means just that. A translation $\delta x$ multiplies by $exp(i\delta x p/\hbar)$ where the p in question is the non-invariant canonical momentum.

-- The non-invariance of the canonical momentum in this case, plus its being the generator of translation, means simply that translation symmetry does not hold and the canonical momentum is not conserved (which we already knew). But is $mv$ conserved?

-- This still leaves me wondering which one is replaced by $-i \hbar \partial_x$, or. more importantly, why.

 

[pines-demon]

That the mechanical momentum traces the trajectory is just because it is proportional to the acceleration.

Well this makes sense, it is the one related to the position of the particle, at least classically.

-- The canonical momentum is the generator of translation means just that. A translation $\delta x$ multiplies by $exp(i\delta x p/\hbar)$ where the p in question is the non-invariant canonical momentum.

Yes.

-- The non-invariance of the canonical momentum in this case, plus its being the generator of translation, means simply that translation symmetry does not hold and the canonical momentum is not conserved (which we already knew). But is $mv$ conserved?

I guess what is conserved is not $m\mathbf v$ but the kinetic energy $mv^2/2$? You do not need quantum mechanics to see that.

-- This still leaves me wondering which one is replaced by $-i \hbar \partial_x$, or. more importantly, why.

The answer to this is clear, it is the canonical momentum that gets transformed into $-i\hbar \nabla$, that's how canonical quantization works and you can check this by just looking examples of calculations, like for the Landau levels. If you use other approaches (like Ballentine's book), the one that becomes $-i\hbar \nabla$ is the one associated to translation invariance, again the canonical momentum.