Feynman's Lectures volume III (Ch:8) -- Resolution of vector states

[Ishika_96_sparkles]

TL;DR Summary

Regarding to the 'resolution of vector states' and the "waiting operator" connecting two states.

In the section 8-2 dealing with resolving the state vectors, we learn that

$$|\phi \rangle =\sum_i C_i | i \rangle$$​

and the dual vector is defined as

$$\langle \chi | =\sum_j D^*_j \langle j |$$​

Then, the an inner product is defined as

$$\langle \chi | \phi \rangle =\sum_{ij} D^*_j C_i \langle j | i \rangle = \sum_{ij} D^*_j C_i \delta_{ij}$$​

This is defined as the amplitude to go from state

$$| \phi \rangle \rightarrow \langle \chi |$$​

later, we encounter the concept of time evolution of the state in sec. 8-3 i.e.

$$\langle \chi |U(t_2,t_1)| \phi \rangle$$​

implying that an operator (time dependent) is involved and is defined as

$$\langle i|U(t_2,t_1)| j \rangle$$​

meaning that this operator links the base states $| j \rangle$ and $\langle i|$ in the time interval $\Delta t= t_2-t_1$.

We are told that the evolution of the state vector is via the explicit time dependence of the coefficients $C_i (t)$ while the basis vectors are constant.

Q1) The previous inner product as defined in the sec. 8-2 had no involvement of the change over time of the state vectors. Then how does one state changes to the other, instantly? Should it be understood in terms of the usual mathematical projection operation? Then in that case, the sentence "...the amplitude to go from $|\phi\rangle$ to $\langle \chi |$..." is just the magnitude/extent of the projection of one state over the other state?

Q2) Does an operator like $\langle i|U(r_2,r_1)| j \rangle$ between the two states make any sense? If not then why not? If yes, then could it be considered some sort of spatial evolution operator? and in that case, what would the spatial variation or the spatial derivative of the Hamiltonian matrix $\frac{d H_{ij}}{d r_2}$ imply? Then, the coefficients will have explicit dependence over the space i.e. $C_i (x)$. [Motivation: ψ(r,t) i.e. the wave-function.]

 

[DrClaude]

Q1) The previous inner product as defined in the sec. 8-2 had no involvement of the change over time of the state vectors.

You simply need to consider the states at a given time $t$.

Then how does one state changes to the other, instantly? Should it be understood in terms of the usual mathematical projection operation? Then in that case, the sentence "...the amplitude to go from $|\phi\rangle$ to $\langle \chi |$..." is just the magnitude/extent of the projection of one state over the other state?

Yes and no. Mathematically, it is simply an inner product, as in regular linear algebra. However, physics adds the concept of measurement. Should a measurement be performed on $| \psi \rangle$, then the probability of finding the system in state $| \chi \rangle$ is $| \langle \chi | \psi \rangle$, provided that $| \chi \rangle$ is a possible outcome of that measurement (an eigenstate of the corresponding measurement operator).

Q2) Does an operator like $\langle i|U(r_2,r_1)| j \rangle$ between the two states make any sense?

What you wrote there is not an operator, but an amplitude. $U(r_2,r_1)$ alone is the operator.

If not then why not? If yes, then could it be considered some sort of spatial evolution operator? and in that case, what would the spatial variation or the spatial derivative of the Hamiltonian matrix $\frac{d H_{ij}}{d r_2}$ imply? Then, the coefficients will have explicit dependence over the space i.e. $C_i (x)$. [Motivation: ψ(r,t) i.e. the wave-function.]

I don't understand what you are getting at here.

 

[Ishika_96_sparkles]

Thank you for your answer.

However, physics adds the concept of measurement. Should a measurement be performed on |ψ⟩, then the probability of finding the system in state |χ⟩ is |⟨χ|ψ⟩| provided that |χ⟩| is a possible outcome of that measurement (an eigenstate of the corresponding measurement operator).

So, is this what it means? If we have a state $|\psi \rangle$ and we take the following inner product
$$\langle \psi| \hat{\mathbf{r}}| \psi \rangle$$
then,we would get the probability to find the system in a certain position Eigenstate $| \mathbf{r} \rangle$, after making a measurement on the state? where, we could write $\langle \psi | \hat{\mathbf{r}}$ as $\langle \chi |$.

I don't understand what you are getting at here.

Since, the wave-function is given by $\psi(r,t)$, then just as there is an operator of time evolution of the state, could we not have something that ``''spatially evolves'' the states across the state space? Like some spatial evolution of the states facilitated by an operator $U(r_2,r_1)$ i.e. in some interval $\Delta r$. This is just an attempt to make an analogy with the time evolution operator. Here, the operator connects two states existing at two different points in state space.

My question is, as to whether this kind of operator is physically meaningful?

The rest of the mathematical relations given are just put in 1-to-1 correspondence with the derivations given in sec. 8-3 for the time evolution operator but with variable 't' replaced by the variable 'r'.

 

[DrClaude]

So, is this what it means? If we have a state $|\psi \rangle$ and we take the following inner product
$$\langle \psi| \hat{\mathbf{r}}| \psi \rangle$$
then,we would get the probability to find the system in a certain position Eigenstate $| \mathbf{r} \rangle$, after making a measurement on the state? where, we could write $\langle \psi | \hat{\mathbf{r}}$ as $\langle \chi |$.

Mathematically, this works, but physically, some of the meaning is lost. The expectation value of an operator over an arbitrary state $| \psi \rangle$ is
$$
\langle O \rangle = \langle \psi | \hat{O} | \psi \rangle
$$
so there is a connection with actual experiments: if you measure $O$ on an ensemble of identical systems prepared in state $| \psi \rangle$, the average of the measurements will be $\langle O \rangle$.

You could define a state $| \chi \rangle \equiv \hat{O} | \psi \rangle$ and then calculate the probability of finding the system in state $| \chi \rangle$, but unless $\psi \rangle$ is an eigenstate of $\hat{O}$, there is no way to make a link with what can be observed in an experiment.

Since, the wave-function is given by $\psi(r,t)$, then just as there is an operator of time evolution of the state, could we not have something that ``''spatially evolves'' the states across the state space? Like some spatial evolution of the states facilitated by an operator $U(r_2,r_1)$ i.e. in some interval $\Delta r$. This is just an attempt to make an analogy with the time evolution operator. Here, the operator connects two states existing at two different points in state space.

My question is, as to whether this kind of operator is physically meaningful?

The rest of the mathematical relations given are just put in 1-to-1 correspondence with the derivations given in sec. 8-3 for the time evolution operator but with variable 't' replaced by the variable 'r'.

To make continuous transformations of the state, you can build unitary operators $\hat{U}(s)$ using the prescription
$$
\hat{U}(s) \equiv e^{i \hat{K} s}
$$
where $\hat{K}$ is a Hermitian operator, known as the generator of the transformation, and $s$ is some continuous parameter. In some cases, these transformations end up having a special significance. Taking the parameter as time $t$, to construct the operator that takes the state from time $t_1$ to $t_2$, it turns out that the generator is $\hat{K} = - \hat{H} / \hbar$, i.e., the Hamiltonian is the generator of time translation, and recover the time evolution operator that you know.

What you are after is the generator of translation in space, which turns out to be
$$
\hat{K} = -\frac{\hat{\mathbf{p}}}{\hbar}
$$
such that the translation operator is
$$
\hat{T}(\Delta \mathbf{r}) \equiv \exp \left[- \frac{i \hat{\mathbf{p}} \cdot \Delta \mathbf{r}}{\hbar} \right],
$$
i.e.,
$$
\hat{T}(\Delta \mathbf{r}) \psi(\mathbf{r}) = \psi(\mathbf{r} + \Delta \mathbf{r})
$$
so momentum is the generator of translation in space.

 

[Ishika_96_sparkles]

I had arrived at a similar conclusion that momentum will come into picture for such an operator.

Now i could see in
$$\hat{T}(\Delta \mathbf{r}) \equiv \exp \left[- \frac{i \hat{\mathbf{p}} \cdot \Delta \mathbf{r}}{\hbar} \right]$$

$\hat{\mathbf{p}} \rightarrow -\iota \hbar\frac{\partial }{\partial r}$ would give such a condition.