I Basic question on meaning of momentum operator

[Kashmir]

How do we apply the momentum operator on a wavefunction?

Wikipedia says
> the momentum operator can be written in the position basis as: ${ }^{[2]}$
$\hat{\mathbf{p}}=-i \hbar \nabla$
where $\nabla$ is the gradient operator, $\hbar$ is the reduced Planck constant, and $i$ is the imaginary unit.
Does this mean that $\hat{P} \psi=-i \hbar \nabla \psi=-i \hbar\left(\hat{i} \frac{\partial}{\partial x} \psi+\hat{\jmath} \frac{\partial}{\partial y} \psi+\hat{k} \frac{\partial}{\partial z} \psi\right)~?$

I'm not sure this is correct

Can anyone please help me

 

[Meir Achuz]

That is the correct definition of $\nabla$.

 

[jambaugh]

The meaning of that operator is that its eigenvalues are the observed values of (the components of) momentum. Likewise, the eigenvectors (eigenfunctions) represent the modes of the system defined with those specified values.

 

[PeterDonis]

I'm not sure this is correct

It's correct. An easy way to verify that is to apply it to a wave function that you know is an eigenfunction of momentum and see what you get.

 

[vanhees71]

It's correct. The question rather is, why this is so.

The answer is Noether's theorem, applied to analytical point-particle mechanics. It says that if there is a symmetry there is also a conserved quantity. The most obvious symmetries are related to the model of space and time. Here we use the Newtonian spacetime model, and one symmetry is translation symmetry, which says that the physics is the same everywhere, i.e., if you take an experiment from one place to an arbitrary other place you will get the same results. The corresponding analysis in classical analytical mechanics shows that the conserved quantity due to this spatial translation invariance is momentum, and thus the momentum operator should "generate translations". That means the operator $\mathrm{i} \hat{\vec{p}} \cdot \delta \vec{a}/\hbar$ should give the change of a wave function under infinitesimal translations in space by $\delta \vec{a}$, i.e.,
$$\psi(\vec{x}+\mathrm{\delta} \vec{a})=\psi(\vec{x})+\delta \vec{a} \cdot \vec{\nabla} \psi(\vec{x}) \stackrel{\text{def}}{=} \psi(\vec{x})+\mathrm{i} \hat{\vec{p}} \cdot \delta \vec{a}/\hbar \psi(\vec{x}),$$
from which you read off
$$\vec{\nabla} = \mathrm{i}/\hbar \hat{\vec{p}} \; \Rightarrow \; \hat{\vec{p}} = -\mathrm{i} \hbar \vec{\nabla}.$$

 

[Kashmir]

I found this expression in my book $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$

The lhs is a scalar because its a bra ket and the rhs then is a vector.

[This post has been lightly edited by the mentors to fix the latex]

 

[PeterDonis]

The lhs is a scalar because its a bra ket and the rhs then is a vector.

No, it isn't, it's actually a vector because the operator in the middle of the bra ket is a vector operator. In other words, this equation is actually shorthand for three equations, for the three corresponding components of the operator $\hat{\mathbf{p}}$ and the operator $\nabla$.

 

[Kashmir]

No, it isn't, it's actually a vector because the operator in the middle of the bra ket is a vector operator. In other words, this equation is actually shorthand for three equations, for the three corresponding components of the operator $\hat{\mathbf{p}}$ and the operator $\nabla$.

I've studied that if an operator acts on a ket it gives a ket. So when the p operator acts on a ket it'll give a ket then that ket is acted by a bra giving a scalar on the lhs

 

[PeterDonis]

I've studied that if an operator acts on a ket it gives a ket.

When a scalar operator acts on a ket it gives a ket. But a vector operator can't operate directly on a ket; an equation in which a vector operator operates on a ket is actually, as I said, shorthand for three equations, one for each component of the vector operator. Each component is a scalar, so in each of the three equations we have a scalar operating on a ket to give a ket, which then gets contracted with a bra to get a scalar (and on the RHS we have one component of the $\nabla$ operator, i.e., an ordinary derivative operator, operating on a scalar to give a scalar).

 

[Kashmir]

When a scalar operator acts on a ket it gives a ket. But a vector operator can't operate directly on a ket; an equation in which a vector operator operates on a ket is actually, as I said, shorthand for three equations, one for each component of the vector operator. Each component is a scalar, so in each of the three equations we have a scalar operating on a ket to give a ket, which then gets contracted with a bra to get a scalar (and on the RHS we have one component of the $\nabla$ operator, i.e., an ordinary derivative operator, operating on a scalar to give a scalar).

So $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle$ is the abbreviation for

$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$
$=-i \hbar\left[e_{x} \frac{\partial}{\partial x}\langle r \mid \psi\rangle+e_{y} \frac{\partial}{\partial y}\langle r \mid \psi\rangle+e_{z} \frac{\partial\langle r \mid \psi\rangle}{\partial z}\right]$ ?

 

[PeterDonis]

So $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle$ is the abbreviation for

$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$
$=-i \hbar\left[e_{x} \frac{\partial}{\partial x}\langle r \mid \psi\rangle+e_{y} \frac{\partial}{\partial y}\langle r \mid \psi\rangle+e_{z} \frac{\partial\langle r \mid \psi\rangle}{\partial z}\right]$ ?

Yes.

 

[Kashmir]

Yes.

Thankyou so much :)

 

[Kashmir]

It's correct. The question rather is, why this is so.

The answer is Noether's theorem, applied to analytical point-particle mechanics. It says that if there is a symmetry there is also a conserved quantity. The most obvious symmetries are related to the model of space and time. Here we use the Newtonian spacetime model, and one symmetry is translation symmetry, which says that the physics is the same everywhere, i.e., if you take an experiment from one place to an arbitrary other place you will get the same results. The corresponding analysis in classical analytical mechanics shows that the conserved quantity due to this spatial translation invariance is momentum, and thus the momentum operator should "generate translations". That means the operator $\mathrm{i} \hat{\vec{p}} \cdot \delta \vec{a}/\hbar$ should give the change of a wave function under infinitesimal translations in space by $\delta \vec{a}$, i.e.,
$$\psi(\vec{x}+\mathrm{\delta} \vec{a})=\psi(\vec{x})+\delta \vec{a} \cdot \vec{\nabla} \psi(\vec{x}) \stackrel{\text{def}}{=} \psi(\vec{x})+\mathrm{i} \hat{\vec{p}} \cdot \delta \vec{a}/\hbar \psi(\vec{x}),$$
from which you read off
$$\vec{\nabla} = \mathrm{i}/\hbar \hat{\vec{p}} \; \Rightarrow \; \hat{\vec{p}} = -\mathrm{i} \hbar \vec{\nabla}.$$

This is a little bit above my level right now. I'll come to it after some time. Thank you :)

 

[Kashmir]

I understood using a translation operator $\hat{T}(\alpha)=e^{-i \alpha \hat{p_x} / \hbar}$ how

$\left\langle x\left|\hat{p}_{x}\right| \psi\right\rangle=-i \hbar \frac{\partial}{\partial x}\langle x \mid \psi\rangle$ and similarly for $p_y$ and $p_z$.

What I don't understand Is the below equation, Is it correct? How can I derive it from the first?

$\left\langle\mathbf{r}\left|\hat{p}_{x}\right| \psi\right\rangle \stackrel{?}{=}-i \hbar \frac{\partial}{\partial x}\langle\mathbf{r} \mid \psi\rangle$

 

[vanhees71]

I found this expression in my book $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$

The lhs is a scalar because its a bra ket and the rhs then is a vector.

[This post has been lightly edited by the mentors to fix the latex]

I guess the confusion is because of the use of bold-face symbols for vectors. I prefer the notation $\hat{\vec{p}}$ to stress that we consider the three operators $\hat{p}_1$, $\hat{p}_2$, and $\hat{p}_3$ as a operator-valued three-component column vector. These denote the abstract representation independent self-adjoint operators acting on kets.

In a slight abuse of notation one uses the same symbol for the operator acting on wave functions, $\psi(\vec{x})=\langle \vec{x}|\psi \rangle$, i.e., one writes
$$\hat{\vec{p}} \psi(\vec{x}) \stackrel{\text{def}}{=} \langle \vec{x}|\hat{p}|\psi \rangle = -\mathrm{i} \hbar \vec{\nabla} \psi(\vec{x}).$$
On the left-hand side it's an operator acting on position-space wave functions, in the middle it's an abstract representation-independent operator acting to the right on $|\psi \rangle$. The last equality follows from the derivation above, using the meaning of the momentum operators as generators of translations in space.

 

[vanhees71]

I understood using a translation operator $\hat{T}(\alpha)=e^{-i \alpha \hat{p_x} / \hbar}$ how

$\left\langle x\left|\hat{p}_{x}\right| \psi\right\rangle=-i \hbar \frac{\partial}{\partial x}\langle x \mid \psi\rangle$ and similarly for $p_y$ and $p_z$.

What I don't understand Is the below equation, Is it correct? How can I derive it from the first?

$\left\langle\mathbf{r}\left|\hat{p}_{x}\right| \psi\right\rangle \stackrel{?}{=}-i \hbar \frac{\partial}{\partial x}\langle\mathbf{r} \mid \psi\rangle$

Yes, it's correct, and the notation with the gradient/nabla operator comes from the definition
$$\hat{\vec{p}}=\begin{pmatrix} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}=-\mathrm{i} \hbar \begin{pmatrix}\partial_x \\ \partial_y \\ \partial_z \end{pmatrix} = -\mathrm{i} \hbar \vec{\nabla}.$$

 

[Kashmir]

Yes, it's correct, and the notation with the gradient/nabla operator comes from the definition
$$\hat{\vec{p}}=\begin{pmatrix} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}=-\mathrm{i} \hbar \begin{pmatrix}\partial_x \\ \partial_y \\ \partial_z \end{pmatrix} = -\mathrm{i} \hbar \vec{\nabla}.$$

Thank you so much.

Can you please tell me how does $\left\langle\mathbf{r}\left|\hat{p}_{x}\right| \psi\right\rangle \stackrel{?}{=}-i \hbar \frac{\partial}{\partial x}\langle\mathbf{r} \mid \psi\rangle$
follow from
$\left\langle x\left|\hat{p}_{x}\right| \psi\right\rangle=-i \hbar \frac{\partial}{\partial x}\langle x \mid \psi\rangle$ ?

 

[vanhees71]

It's, because formally $|\vec{r} \rangle \equiv |x,y,z \rangle=|x \rangle \otimes |y \rangle \otimes |z \rangle$.

 

[jambaugh]

It is important to understand we're in two distinct spaces with distinct meanings. There's physical 3-dimensional space and the Hilbert space of wave functions represented by the "ket"s... (and also there's the dual space of "bra"s). I think it would be best to begin working in one (spatial) dimension while trying to understand the deBroglie relation $\hat{p} = -i\hbar \frac{d}{dx}$ and then when that's soundly understood, the 3-dimensional and 3+1 dimensional generalizations are no problem.

It is also, I believe, useful to think about the linear algebraic "What's going on" as one tries to understand the "why". What we have here is a Lie group of system transformations (not necessarily symmetries) which include translations of position and translations of momentum, and also one might add spatial rotations. These are all things that can happen to a particle as it evolves over time. Associated with that group is its Lie algebra of generators. The exponential map relates them via $g(t) = \exp(t\Gamma)$.

Finally, there is a representation of this group and its Lie algebra that defines our system. Classical systems are represented by points in phase space (or more generally, allowing probabilistic descriptions, a distribution over phase space. The system transformations are expressed as a class of diffeomorphisms (flows) in phase space we call canonical transformations. They are the ones that preserve its symplectic structure. So in that representation, the classical correspondent to de Broglie's relation is (double-check my choice of conventions):
\hat{p}[f(p,x)] = \{f,p\} = \frac{\partial f}{\partial x}\frac{\partial p}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial p}{\partial x} = \frac{\partial}{\partial x} f(p,x)
In short, classically $\hat{p} = \frac{\partial}{\partial x}$, but this is a contextual equality with an implicit "in this representation". Likewise $\hat{x} = -\frac{\partial }{\partial p}$

Note that from a linear algebraic point of view I could name the phase space variables anything, e.g. $(a,b)$ in which case $\hat{p}=\frac{\partial}{\partial b}$, and $\hat{x} = -\frac{\partial}{\partial a}$. We're only talking about a mathematical representation at this point and the associations with the labels come from the physics.

There's a lot more to be said here but this post is already too long. I'll follow up later to address more how the physics dictates the mathematics.

 

[jambaugh]

Now I would address the physics we express in the algebraic framework I reference in the prior post.

I personally believe that Noether's work in the form of her theorem is one of the most significant results in the history of mathematical physics. There is more we can infer than the mere theorem itself. Noether formally proved something that was already an observed phenomenon, that associated with every continuous symmetry there was a corresponding conserved quantity. But here is where I believe it goes even further.

Consider that any observable must be repeatedly observable... it thus must be a quantity that is conservable i.e. we can, in principle, within the laboratory, set up the dynamics of a given system so that any given observable quantity is conserved. I thus assert that we can extend Noether's theorem not just to existing symmetries and conserved quantities but, in fact, to any potential symmetries i.e. any continuous system transformation, and to any conservable quantities i.e. any physical observables. While I've not, in my limited reading, seen it expressed this way I think it is implicit in the framework of modern physics.

Thus, and in particular here, we associate with any observable a corresponding generator of a system transformation, $p \sim \hat{P}$, $x\sim \hat{X}$, $\ell_z\sim \hat{L}_z$, etc. The fundamental language of modern physics is that of Lie groups/algebras and their representations. In classical mechanics these are point representations; however, that is only for statistically certain cases. We can easily generalize to (probability) distributions of points in which case the representations become linear representations over a "function" space.

One of the failings of classical mechanics is in the definition of entropy along with the pathologies of the continuum. (Consider for example the Banach-Tarski paradox). In classical mechanics, we can, in principle encode the sum total of all human knowledge into the exact position (or momentum) of one particle. Not only is the information content of a single particle infinite, but it is also transfinite i.e. $\aleph_1$ in size.

But, justifications aside, if we consider an alternative theory, our first-next look might be at finite-dimensional or at worst discrete representations of these fundamental Lie groups/algebras of conservables and their corresponding potential symmetries.

I want to argue that quantum theory is the only alternative but I would avoid such hubris that assumes the alternatives I can't imagine are therefore invalid. Let it suffice to say that quantum theory is one alternative and its structure in this context is of finite or at-the-very-least discrete representations of the corresponding group/algebras. Its form is that the maximally specified systems are projectively represented by elements of some vector space, specifically a Hilbert space where the metric structure relates to Born's probability interpretation. The quantity observed is expressed through the eigen-value principle. We can express those vectors as "wave functions" but, in the immortal words of Monty Python "It's only a model". In the end, and in the most general setting, the system states are linear "co-operators" mapping the observable's corresponding operators (Lie algebra elements) to the expected values of the observables for a given system "state". (That is to say density (co)operators.)

The rest of what I could include here is a long and deep study of quantum theory. I think about trying to summarize but it is just too much to include in a single post, or even a single text or body of work by a single individual. That is the job of you students of quantum theory.

So, bringing it back to the OP, your "why" is a very very deep question and I'm not sure how deep you want to go. But the most terse summary I can give you is: The eigen-value principle, The correspondence principle, and The fundamental structure of representations of Lie Groups/Algebras along with Noether's theorem(with my addendum).

 

[Kashmir]

Yes, it's correct, and the notation with the gradient/nabla operator comes from the definition
$$\hat{\vec{p}}=\begin{pmatrix} \hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}=-\mathrm{i} \hbar \begin{pmatrix}\partial_x \\ \partial_y \\ \partial_z \end{pmatrix} = -\mathrm{i} \hbar \vec{\nabla}.$$

so is this correct $\overrightarrow{\hat{p}}|\psi\rangle=\left(\begin{array}{l}P_{x}|\psi\rangle \\ P_{y}|\psi\rangle \\ P_{z}|\psi\rangle\end{array}\right)$

 

[vanhees71]

Yes, that's the definition of the operator $\hat{\vec{p}}$. The $P_j$ are of course also operators, i.e., you should write $\hat{P}_j$ (with $j \in \{x,y,z \}$).

 

[PeroK]

I found this expression in my book $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$

The lhs is a scalar because its a bra ket and the rhs then is a vector.

This is essentially the 3D version of the question you posted last month:

https://www.physicsforums.com/threads/clarification-on-how-bra-ket-works-here.1012106/#post-6597505

There are several ways to work in 3D. Let's take Newton's second law as an example. We have:
$$F_x = ma_x, \ F_y = ma_y, \ F_z = ma_z$$And, you simply work with three separate equations.

This, however, is completely equivalent to using vector notation to express those three equations as one vector equation:$$ \vec F = m \vec a$$Or, in using boldface $$\mathbf F = m \mathbf a$$Or, of course, you could write the components in matrix notation:
$$
\begin{bmatrix}
F_x \\
F_y \\
F_z
\end{bmatrix} =
m
\begin{bmatrix}
a_x \\
a_y \\
a_z
\end{bmatrix}
$$These four formulations are all completely equivalent. In your case, what you have is equivalent to:
$$\langle x|\hat p_x |\psi \rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$$$$\langle y|\hat p_y |\psi \rangle = -i\hbar\frac{\partial}{\partial y}\langle y|\psi\rangle$$$$\langle z|\hat p_z |\psi \rangle = -i\hbar\frac{\partial}{\partial z}\langle z|\psi\rangle$$

 

[Kashmir]

Yes, that's the definition of the operator $\hat{\vec{p}}$. The $P_j$ are of course also operators, i.e., you should write $\hat{P}_j$ (with $j \in \{x,y,z \}$).

So it's also equivalent to write $\begin{aligned} \hat{p}|\psi\rangle &=e_{x} \hat{p}_{x}|\psi\rangle \\ &+e_{y} \hat{p}_{y}|\psi\rangle \\ &+e_{z} \hat{p}_{z}|\psi\rangle \end{aligned}$

Where $e_{x}$ is the unit vector in x direction.

This expression isn't a sum of kets but a sum of kets with spatial unit vector coeffiencts.

I know that a bra can act on a ket however it's first time I'm seeing a bra acted on such a ket with spatial unit vector coefficients.
So is it that a bra of this expression i.e $\langle r|\hat{p}| \psi\rangle$
is just defined to mean

$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$?

 

[Kashmir]

This is essentially the 3D version of the question you posted last month:

https://www.physicsforums.com/threads/clarification-on-how-bra-ket-works-here.1012106/#post-6597505

There are several ways to work in 3D. Let's take Newton's second law as an example. We have:
$$F_x = ma_x, \ F_y = ma_y, \ F_z = ma_z$$And, you simply work with three separate equations.

This, however, is completely equivalent to using vector notation to express those three equations as one vector equation:$$ \vec F = m \vec a$$Or, in using boldface $$\mathbf F = m \mathbf a$$Or, of course, you could write the components in matrix notation:
$$
\begin{bmatrix}
F_x \\
F_y \\
F_z
\end{bmatrix} =
m
\begin{bmatrix}
a_x \\
a_y \\
a_z
\end{bmatrix}
$$These four formulations are all completely equivalent. In your case, what you have is equivalent to:
$$\langle x|\hat p_x |\psi \rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$$$$\langle y|\hat p_y |\psi \rangle = -i\hbar\frac{\partial}{\partial y}\langle y|\psi\rangle$$$$\langle z|\hat p_z |\psi \rangle = -i\hbar\frac{\partial}{\partial z}\langle z|\psi\rangle$$

Thank you. But adding a spatial unit vector coefficient to a ket is very new to me. Please see the reply above

 

[PeroK]

I know that a bra can act on a ket however it's first time I'm seeing a bra acted on such a ket with spatial unit vector coefficients.

A bra or ket representing position must have three components, right? You may start learning QM with the simpler 1D case, but sooner or later you have to extend the theory to three spatial dimensions. That shouldn't some as a surprise!

 

[vanhees71]

It's rather a product. The common eigenvector of the three compatible position-vector components is formally a Kronecker product, $|x,y,z \rangle=|x \rangle \otimes |y \rangle \otimes |z \rangle$. By definition
$$\hat{\vec{x}} |x,y,z \rangle = \begin{pmatrix} x \\ y \\ z \end{pmatrix} |x,y,z \rangle$$
is simply a useful notation. There's nothing problematic with it. I still don't understand, what the problem should be.

 

[PeroK]

Thank you. But adding a spatial unit vector coefficient to a ket is very new to me. Please see the reply above

PS as noted in some of the posts above, it's not correct to use Euclidean 3D unit vectors with states/vectors in Hilbert space. The states represent position in 3D space, but are themselves not vectors in Euclidean 3D space.

So, actually, you can't push the analogy I gave in post #23 too far.

 

[Kashmir]

PS as noted in some of the posts above, it's not correct to use Euclidean 3D unit vectors with states/vectors in Hilbert space. The states represent position in 3D space, but are themselves not vectors in Euclidean 3D space.

So, actually, you can't push the analogy I gave in post #23 too far.

That's why I'm having trouble understanding the generalisation to three dimensions.

I know so many people have explained it that I'm ashamed to ask again.

1) First of all, is a spatial unit vector coefficient to a ket a valid thing to write ? i.e $e_{x}|\psi\rangle$

2)as Mr vanhees explained we can write
$\overrightarrow{\hat{p}}|\psi\rangle=\left(\begin{array}{l}P_{x}|\psi\rangle \\ P_{y}|\psi\rangle \\ P_{z}|\psi\rangle\end{array}\right)$

so can we write it as :$\begin{aligned} \hat{p}|\psi\rangle &=e_{x} \hat{p}_{x}|\psi\rangle \\ &+e_{y} \hat{p}_{y}|\psi\rangle \\ &+e_{z} \hat{p}_{z}|\psi\rangle \end{aligned}$

3) If we can write it as above then can I apply a bra on this expression, if I can then what will happen if I apply a bra $\langle r$ on the above expression?

Is it going to be written as:
$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$ ?

 

[PeroK]

Is it going to be written as:
$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$ ?

No, that's wrong. The correct formalism was shown in post #27:

It's rather a product. The common eigenvector of the three compatible position-vector components is formally a Kronecker product, $|x,y,z \rangle=|x \rangle \otimes |y \rangle \otimes |z \rangle$. By definition
$$\hat{\vec{x}} |x,y,z \rangle = \begin{pmatrix} x \\ y \\ z \end{pmatrix} |x,y,z \rangle$$
is simply a useful notation. There's nothing problematic with it. I still don't understand, what the problem should be.

 

[PeroK]

Is it going to be written as:
$e_{x}\left\langle r\left|P_{x}\right| \psi\right\rangle+e_{y}\left\langle r\left|P_{y}\right| \psi\right\rangle+e_{z}\left\langle r\left|P_{z}\right| \psi\right\rangle$ ?

Note that in general $\langle \phi|\hat X |\psi \rangle$ is a complex number!

 

[Kashmir]

Also $e_{x}|\psi\rangle$ is wrong?

 

[PeroK]

Also $e_{x}|\psi\rangle$ is wrong?

You have a unit vector in Euclidean space with a state vector in an abstract Hilbert space. That's not a valid combination.

 

[Kashmir]

So the correct expression is $\langle r|\overrightarrow{\hat{P}}| \psi\rangle=\left(\begin{array}{c}\left\langle r\left|P_{x}\right| \psi\right\rangle \\ \left\langle r\left|P_{y}\right| \psi\right\rangle \\ \left\langle r\left|P_{z}\right| \psi\right.\end{array}\right)$

 

[PeroK]

So the correct expression is $\langle r|\overrightarrow{\hat{P}}| \psi\rangle=\left(\begin{array}{c}\left\langle r\left|P_{x}\right| \psi\right\rangle \\ \left\langle r\left|P_{y}\right| \psi\right\rangle \\ \left\langle r\left|P_{z}\right| \psi\right.\end{array}\right)$

There's no single "correct" expression. That's a shorthand for the full "Tensor Product" formalism.

 

[Kashmir]

There's no single "correct" expression. That's a shorthand for the full "Tensor Product" formalism.

I've started learning tensor product formalism.

For a particle which can move in three dimensions is it's momentum operator :
$\hat{p}=\left(\hat{p}_{x} \otimes 1 \otimes 1\right)+\left(1 \otimes \hat{p}_{y} \otimes 1\right)+\left(1 \otimes 1 \otimes \hat{p}_{z}\right)$ ?

 

[vanhees71]

Formally yes, but you'll hardly find any textbook, writing it in this way.

 

[Kashmir]

Formally yes, but you'll hardly find any textbook, writing it in this way.

Then this big operator acts on a ket which is itself a tensor product of the basis states of x, y and z.

 

[vanhees71]

You can interpret the Hilbert space of a spin-0 particle as the product space spanned by the (generalized) eigenvectors of the position-vector-component operators $\hat{x}$, $\hat{y}$, and $\hat{z}$, i.e., a general vector in this space
$$|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \int_{\mathbb{R}} \mathrm{d} z |x \rangle \otimes |y \rangle \otimes |z \rangle.$$
Usually you write for better readability
$$|x,y,z \rangle=|x \rangle \otimes |y \rangle \otimes |z \rangle.$$

 

[Kashmir]

You can interpret the Hilbert space of a spin-0 particle as the product space spanned by the (generalized) eigenvectors of the position-vector-component operators $\hat{x}$, $\hat{y}$, and $\hat{z}$, i.e., a general vector in this space
$$|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \int_{\mathbb{R}} \mathrm{d} z |x \rangle \otimes |y \rangle \otimes |z \rangle.$$
Usually you write for better readability
$$|x,y,z \rangle=|x \rangle \otimes |y \rangle \otimes |z \rangle.$$

Thank you.
I'm trying to expand
$\left\langle r\left|p\right| \Psi\right\rangle$ by tensor product and am stuck how to evaluate the below integral
$\int d x^{\prime} \psi(x', y', z') \frac{\partial}{\partial x} \delta\left(x-x^{\prime}\right)$

Below is the expression i have among other terms while expanding $\left\langle r\left|p_{x}\right| \Psi\right\rangle$

$\int d x^{\prime} \psi\left(x^{\prime}, y, z\right)\left\langle x|P | x^{\prime}\right\rangle$

Any hints please?

 

[vanhees71]

Hint: $\partial_x \delta(x-x')=-\partial_{x'} \delta(x-x')$.

 

[Kashmir]

$\hat{P}=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes 1\right)+\left(\mathbb{1} \otimes \hat{P}_{y} \otimes \mathbb{1}\right)+\left(\mathbb{1} \otimes \mathbb{1} \otimes \hat{p}_{2}\right)$
Then
$\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$ where ... are for similar two terms. If we just work out the first term we can write it as

$\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$$=\left.\iiint d x^{\prime} d y^{\prime} d z^{\prime} \hat{P}_{x}\left|x^{\prime}\right\rangle \otimes\left|y^{\prime}>\otimes\right| z^{\prime}\right\rangle \psi\left(x^{\prime}, y^{\prime}, z\right)$

Then $\langle r|\hat{p}| \psi\rangle=$$\langle x|\otimes<y| \otimes\langle z|$$\iiint d x^{\prime} d y^{\prime} d z^{\prime} \hat{p}_{x}\left|x^{\prime}\right\rangle \otimes\left|y^{\prime}\right\rangle \otimes\left|z^{\prime}\right\rangle \psi\left(x 'y 'z' ^{\prime}\right)$

$=\iiint d x' d y' d z'^{\prime}\left\langle x\left|\hat{P}_{x}\right| x^{\prime}\right\rangle \cdot \delta\left(y-y^{\prime}\right) \cdot \delta\left(z-z^{\prime}\right) \psi(x' y'z')$

$=\int d x^{\prime}\left\langle x\left|\hat{p}_{x}\right| x'\right\rangle \psi(x, y, z)$I don't know how to simplify it further.

 

[Kashmir]

$\hat{P}=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes 1\right)+\left(\mathbb{1} \otimes \hat{P}_{y} \otimes \mathbb{1}\right)+\left(\mathbb{1} \otimes \mathbb{1} \otimes \hat{p}_{2}\right)$
Then
$\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$ where ... are for similar two terms. If we just work out the first term we can write it as

$\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$$=\left.\iiint d x^{\prime} d y^{\prime} d z^{\prime} \hat{P}_{x}\left|x^{\prime}\right\rangle \otimes\left|y^{\prime}>\otimes\right| z^{\prime}\right\rangle \psi\left(x^{\prime}, y^{\prime}, z\right)$

Then $\langle r|\hat{p}| \psi\rangle=$$\langle x|\otimes<y| \otimes\langle z|$$\iiint d x^{\prime} d y^{\prime} d z^{\prime} \hat{p}_{x}\left|x^{\prime}\right\rangle \otimes\left|y^{\prime}\right\rangle \otimes\left|z^{\prime}\right\rangle \psi\left(x 'y 'z' ^{\prime}\right)$

$=\iiint d x' d y' d z'^{\prime}\left\langle x\left|\hat{P}_{x}\right| x^{\prime}\right\rangle \cdot \delta\left(y-y^{\prime}\right) \cdot \delta\left(z-z^{\prime}\right) \psi(x' y'z')$

$=\int d x^{\prime}\left\langle x\left|\hat{p}_{x}\right| x'\right\rangle \psi(x, y, z)$I don't know how to simplify it further.

We can write it further as :$\int d x^{\prime}\left(\frac{\hbar}{i} \frac{\partial}{\partial x} \delta\left(x-x^{\prime}\right)\right) \psi\left(x^{\prime}, y, z\right)$

$=-\frac{\hbar}{i} \int d x^{\prime}\left(\frac{\partial}{\partial x'} \delta\left(x-x^{\prime}\right)\right) \psi\left(x^{\prime}, y, z\right)$

$=-\frac{\hbar}{i}\left[\left.\psi(x', y, z) \delta(x-x')\right|_{-\infty} ^{\infty}-\int\left(\frac{d}{d x'} \psi(x', y, z)\right) \delta(x-x') d x'\right]$

$=\frac{\hbar}{i}\left[\frac{d}{d x} \psi(x, y, z)\right]$

Is that correct?

 

[Kashmir]

There's no single "correct" expression. That's a shorthand for the full "Tensor Product" formalism.

Using the tensor product formalism in which $\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$
I found $\langle r|\hat{P}| \psi\rangle=\left(\frac{\hbar}{i} \frac{\partial}{\partial x} \psi+\frac{\hbar}{i} \frac{\partial}{\partial y} \psi+\frac{\hbar}{i} \frac{\partial}{\partial z} \psi\right).$ . This doesn't have any spatial unit vectors in it.

But in *Townsends Quantum mechanics* he says that $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$. This has spatial unit vector.

How to reconcile the two now?

 

[PeroK]

Using the tensor product formalism in which $\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$
I found $\langle r|\hat{P}| \psi\rangle=\left(\frac{\hbar}{i} \frac{\partial}{\partial x} \psi+\frac{\hbar}{i} \frac{\partial}{\partial y} \psi+\frac{\hbar}{i} \frac{\partial}{\partial z} \psi\right).$ . This doesn't have any spatial unit vectors in it.

But in *Townsends Quantum mechanics* he says that $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$. This has spatial unit vector.

How to reconcile the two now?

Let's start with what we know. We have an abstract state $|\psi \rangle$. We know this has a position space representation as a wave-function of three real variables (equivalent to one 3D Euclidean vector). So:
$$ |\psi \rangle \leftrightarrow \psi(x, y, z) \equiv \psi(\vec r)$$Here $\vec r$ is a simple 3D Euclidean vector. Now, we know (or suspect) that the momentum operator acting on a position space wave-function takes the form of a gradient:
$$\hat p |\psi \rangle \leftrightarrow -i \hbar \nabla \psi(x, y, z)$$There are two immediate issues. First, the wave-function is compex valued, so the RHS is not a vector in 3D Euclidean space. It's some vector with three components. You could introduce unit vectors $e_x, e_y, e_z$ but the whole construction is not a simple 3D real Euclidean vector.

The second issue is that the LHS must have three components. There must be some way of representing the Hilbert space so that it is associated with the 3D nature of the gradient. One idea is that we have three compatible position operators $\hat x, \hat y, \hat z$, which have eigenstates $| x \rangle, |y \rangle, |z \rangle$ that together form an uncountable basis for our Hilbert space. Informally we can combine these into one "vector" position eigenstate:
$$\mathbf r \equiv |x, y, z \rangle$$Note that $\mathbf r$ is not really a vector here, it's just a convenient shorthand. Many QM books will continue under the assumption that we don't need to delve any deeper into the structure of Hilbert spaces. If you do delve deeper, then you need the concept of a tensor product of Hilbert spaces and operators. When the dust settles you end up with:
$$ \hat p |\psi \rangle \equiv \hat p_x \otimes \hat p_y \otimes \hat p_z |\psi \rangle \leftrightarrow -i \hbar \nabla \psi(x, y, z)$$Where $|\psi \rangle$ can be expressed as a tensor product as above:
$$|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \int_{\mathbb{R}} \mathrm{d} z \ \psi(x, y, z) |x \rangle \otimes |y \rangle \otimes |z \rangle.$$And everyting should make sense. We see that the position-space wave-function comes from the coefficients of the state in the position basis. And the momentum operator acting on that corresponds to the gradient of the position space wave-function.

The final step is to take the inner product with the bra $\langle \mathbf r |$ to get the "equality":
$$\langle \mathbf r |\hat p |\psi \rangle = -i \hbar \nabla \psi(x, y, z)$$And, we can see that this really means:
$$\langle x |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle y |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle z |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$And you have to accept that this is what Townsend means. Note that I chose not to represent the momentum operator as a bold-face vector. That highlights that we have to take these equations as symbolic. You can't take the first equation too literally. If you try to add Euclidean basis vectors, then you are pushing the correspondence too far.

 

[PeroK]

To summarise. This is what comes out of the formalism:
$$\langle x |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle y |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle z |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$You can choose how to represent that more compactly according to your taste. I went for:
$$\langle \mathbf r |\hat p |\psi \rangle = -i \hbar \nabla \psi(x, y, z)$$Townsend went for:
$$\langle \mathbf r |\hat {\mathbf p} |\psi \rangle = -i \hbar \nabla \langle \mathbf r |\psi \rangle$$

 

[vanhees71]

Using the tensor product formalism in which $\hat{P}|\psi\rangle=\left(\hat{P}_{x} \otimes \mathbb{1} \otimes \mathbb{1}\right)|\psi\rangle+\ldots$
I found $\langle r|\hat{P}| \psi\rangle=\left(\frac{\hbar}{i} \frac{\partial}{\partial x} \psi+\frac{\hbar}{i} \frac{\partial}{\partial y} \psi+\frac{\hbar}{i} \frac{\partial}{\partial z} \psi\right).$ . This doesn't have any spatial unit vectors in it.

But in *Townsends Quantum mechanics* he says that $\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$. This has spatial unit vector.

How to reconcile the two now?

I don't know what your symbol $\hat{P}$ means. You have three momentum-component operators. In the product notation it's defined as
$$\hat{P}_x=\hat{p}_x \otimes \hat{1} \otimes \hat{1}, \quad \hat{P}_y=\hat{1} \otimes \hat{p}_y \otimes \hat{1}, \quad \hat{P}_z=\hat{1} \otimes \hat{1} \otimes \hat{p}_z.$$
The momentum operators in the position representation are $\hat{p}_j=-\mathrm{i} \hbar \partial/\partial x_j$. This you can derive from the commutation relations $[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk} \hat{1}$ with $j,k \in \{x,y,z \}$.

 

[Kashmir]

...And, we can see that this really means:
$$\langle x |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle y |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle z |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$

Shouldn't this have been $$\langle r|\hat p_x |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle r |\hat p_y |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle r |\hat p_z |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$ ?

 

[PeroK]

Shouldn't this have been $$\langle r|\hat p_x |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle r |\hat p_y |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle r |\hat p_z |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$ ?

I left that as an exercise for you to see why it's the same thing.

 

[Kashmir]

To summarise. This is what comes out of the formalism:
$$\langle x |\hat p |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$

Since $\langle x |$ needs to be in the tensor product space just like
$\langle r|=\langle x| \otimes\langle y| \otimes\langle z|$

What is $\langle x|$ here in the lhs of above quoted equation?