Doubts regarding Quantum Mechanics [closed]

Question

I recently started with Quantum mechanics from The Principles of Quantum Mechanics by R. Shankar. I have studied Linear Algebra in my course already. I have also studied QM from D. J Griffiths(up to finite potential well). I felt the lack of rigor in Griffiths so I switched to Shankar. Over time, I came up with some problems I encountered and would like to clarify them.

• What is the advantage of having a hermitian operator in QM? What is the most useful part of having a hermitian operator?
• How does an operator act on a $|\psi \rangle$ in terms of measurement of the system?
• Using the Fourier transform, we know that:$$\psi(p)=\int_{-\infty}^{\infty}{e^{-ikx}\psi(x)dx}$$So does $|\psi(p)|^2$ tells us about the probability of having the momentum p at time of measurement? If true, can we generalize it to other operators as well?
• For Example,There is a normalized wavefunction $\psi(x)$, does each x describes the state of particle?If yes then a particle has an infinite number of possible states(as it is in an infinite dimensional vector space.).If not, Then what is meant by "State" in Quantum mechanics?
• If I apply momentum operator on the ket $|\psi \rangle$ then,$$P|\psi \rangle = |\psi' \rangle$$ then what does $|\psi'(x)|^2$ represents?What is the difference between $\psi'(x)$ and $\psi(p)$?
  It will be helpful if you can give me some tips on Shankar's book as well as some of you would have studied it.

Answer 1

What is the advantage of having a hermitian operator in QM? What is the most useful part of having a hermitian operator?

Having hermitian (more accurately, self-adjoint) operators is not a matter of convenience; they play a fundamental role in the standard formulation of quantum mechanics because they can represent physical observables. The reason physical observables are modeled with self-adjoint operators is tied to the spectral theorem; the main ideas are (i) that the possible measurement outcomes should be real-valued, and (ii) that for any state vector $\psi$ in the Hilbert space, we can compute a probability distribution over the possible measurement outcomes.

How does an operator act on a $|\psi \rangle$ in terms of measurement of the system?

Measurements are not modeled by the actions of operators on state vectors.

[...] So does $|\psi(p)|^2$ tells us about the probability of having the momentum p at time of measurement? If true, can we generalize it to other operators as well?

Yes to both questions.

For Example,There is a normalized wavefunction $\psi(x)$, does each x describes the state of particle?If yes then a particle has an infinite number of possible states(as it is in an infinite dimensional vector space.).If not, Then what is meant by "State" in Quantum mechanics?

The state of the particle is described by the entire function $\psi$, not the specific value of $\psi(x)$ for some $x$. In QM, a pure state can be represented by a normalized element of the Hilbert space (in this case, a normalized function $\psi$) with the understanding that two unit vectors actually represent the same state if they differ by some overall phase, i.e. $\psi$ and $\phi$ are equivalent if $\psi = e^{i\alpha} \phi$ for some $\alpha\in\mathbb R$.

If I apply momentum operator on the ket $|\psi \rangle$ then,$$P|\psi \rangle = |\psi' \rangle$$ then what does $|\psi'(x)|^2$ represents?What is the difference between $\psi'(x)$ and $\psi(p)$?

$\psi'$ does not represent anything in particular, and has essentially nothing to do with $\psi(p)$. For what it's worth, I would highly suggest you not use the letter $\psi$ for all three of those contexts; quantum mechanics is confusing enough without using the same symbol for three different things at the same time.

Answer 2

For more understanding, you should learn a bit of Sturm-Liouville theory of differential equations to understand why we care about such things.

What is the advantage of having a hermitian operator in QM? What is the most useful part of having a hermitian operator?

When initially trying to build a new theory, it is best to be artificially excessively restrictive so as to be guaranteed good results. It is known before von Neumann formulated modern quantum theory in the current mathematical formalism, that Hermitian operators always have

1. complete set of eigenfunctions suitable to expand the entire Hilbert space
2. real valued eigenvalues so that all measurements immediately make physical sense

It turns out that we do not need to expand beyond Hermitian operators to work out all the consequences of quantum theory, and so we do not. Note that we actually do have known situations whereby entertaining non-Hermitian quantum mechanics is more convenient, but that is a niche thing, and we do not have to entertain the philosophical headaches that that would bring to us.

How does an operator act on a $\left|\psi\right>$ in terms of measurement of the system?

The result of a measurement is literally the content of a postulate.

So does $\left|\psi(\vec p)\right|^2$ tells us about the probability of having the momentum $\vec p$ at time of measurement? If true, can we generalize it to other operators as well?

Yes and yes.

For example, there is a normalized wavefunction $\psi(x)$, does each $x$ describes the state of particle? [...] If not, then what is meant by "state" in Quantum mechanics?

No. $x$ is the position coördinate of spacetime. I suppose you would like to consider infinitely localised states $\psi_{x_0}(x)=\delta(x-x_0)$ as a state, in which case you would have your infinite set of states to expand over, one for every $x_0$. However, this is not what is being meant.

Instead, the entire $\psi(x)$ itself is the (pure) state of quantum particle. Now, to be even more pedantic, $\psi(x)=\left<x|\psi\right>$, and it is $\left|\psi\right>$ that is the (pure) state of the quantum particle, whereas we pretend that $\left<x\right|$ to be a suitable position coördinate basis expansion for that state.

Note that a state would define a probability amplitude distribution over the entirety of space at once.

what does $\left|\psi^\prime(x)\right|^2$ represents?

Note that when $\psi(x)$ is normalised properly, then the derivative is not normalised. So this would not be a probability; at most it is a weighted, scaled, proportional thing to the probability. With a bit of thought, it is the position probability distribution multiplied by square-momentum so that it gives the weight of how much momentum squared is at which position in space. If you integrate this over all positions, then you get the expectation square-momentum value of the state. It is of use in writing down the kinetic energy of a state, in the non-relativistic approximation.

Answer 3

A measurement is just an interaction that copies or transfers information from one system to another. If you're going to use that information to describe experimental results the laws of physics must allow you to copy it. You can work out what kinds of states can produce information that can be copied perfectly and it turns out that those states form an orthogonal set, see Section I of

https://arxiv.org/abs/0707.2832

There is no such thing as the probability of a particle having a particular value of $p$ or $x$ because those are continuous observables and a delta function isn't a physically possible state. Those functions are used to get expectation values by integrating them over a range of values. The physically possible states are smooth functions that may be highly peaked near specific values of position and momentum compared to the scale of everyday life as a result of decoherence:

https://arxiv.org/abs/1111.2189