Basic question on meaning of momentum operator

[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?

 

[Kashmir]

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 $j,k \in \{x,y,z \}$.

I meant this $\hat{p}=\left(\hat{p}_{x} \otimes 1 \otimes 1\right)+\left(1 \otimes \hat{p}_{y} \otimes 1\right)+\left(1 \otimes 1 \otimes \hat{p}_{z}\right)$

 

[PeroK]

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?

I think I was a bit sloppy in using $\langle x|$ to pick out the first component in the tensor product. I think your suggestion to use $\hat p_x$ instead is better and makes more sense.

 

[PeroK]

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

Yes, I think that's better. The key point is that an inner product produces a single complex number (for any Hilbert space). The underlying equality must, therefore, be three separate equalities, that we want to write in more compact notation.

Any vectorization of these three equalities is, therefore, a symbolic/notational thing.

 

[Kashmir]

Thank you.

In the position representation, Hamiltonian of a free particle in 3D is usually given as :
$$-\frac{h^{2}}{2 m} \nabla^{2}$$

To prove this, is the below correct :

$$
\hat{H}=\frac{\hat{p}^{2}}{2 m}=\frac{1}{2 m} \hat{p} \cdot \hat{p}
$$

Then the position representation is

$$
\left\langle r\left|\frac{1}{2 m} \hat{p} \cdot \hat{p}\right| \psi\right\rangle
$$

Where we note that
$$
\langle r|=\langle x| \otimes\langle y| \otimes\langle z|
$$ and $$\hat{p}=\left(\hat{p}_{x} \otimes 1 \otimes 1\right)+\left(1 \otimes \hat{p}_{y} \otimes 1\right)+\left(1 \otimes 1 \otimes \hat{p}_{z}\right)$$ and solve this

 

[vanhees71]

The last equation is of course not correct. It doesn't make any sense, or what do you think adding the three components of the momentum wrt. some arbitrary cartesian basis should physically mean? What makes sense is the notation
$$\hat{\vec{p}}=\begin{pmatrix}\hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}.$$
It is the quantum version of the same mathematical description of the momentum components in classical mechanics you also usually put in a column-vector notation.

Than it's also immediately clear that, analogous to classical mechanics (and in this case even without any "operator-ordering problem", because the components of the momentum operator all commute)
$$\hat{\vec{p}}^2=\hat{p}_x^2+\hat{p}_y^2 + \hat{p}_z^2.$$

 

[Kashmir]

How did you go from this

$$\hat{\vec{p}}=\begin{pmatrix}\hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}.$$

To this?

$$\hat{\vec{p}}^2=\hat{p}_x^2+\hat{p}_y^2 + \hat{p}_z^2.$$

It's like you treated it like a euclidean vector

 

[andresB]

It's like you treated it like a euclidean vector

The momentum operator (and the position) are vector operators . Loosely speaking, they have components along each axis. Under a rotation, the components behave/transform pretty much like the components of a Euclidean vectors.

In this case, the $P^{2}$ operator is invariant under rotations, just like the magnitude of an euclidean vector is invariant under rotations. See Messiah ch XIII $10