Very basic questions about operators in QM

[olgerm]

Could you tell me if I have understood following about operators in QM correctly?

Wavefunction takes all generalized coordinates of the system as arguments.
for example if we have a system of proton and electron (in 3-dimensional space) then the wavefunction of this system has 7 arguments $\psi_{example}(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})$. in this example generalized coordinates of the system are normal coordinates of pointbodies of the system.

1. $t$ is time
2. $x_{0.1}$ is projection of position of proton to 1.-coordinate-axis.
3. $x_{0.2}$ is projection of position of proton to 2.-coordinate-axis.
4. $x_{0.3}$ is projection of position of proton to 3.-coordinate-axis.
5. $x_{1.1}$ is projection of position of electron to 1.-coordinate-axis.
6. $x_{1.2}$ is projection of position of electron to 2.-coordinate-axis.
7. $x_{1.3}$ is projection of position of electron to 3.-coordinate-axis.

Probability-density of system having the given generalized coordinates is square of absolute value of the wavefunction with the generalized coordinates given as arguments. $P(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})=|\psi(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})|^2=\psi(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})*complexconjugate(\psi(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3}))$
In example-system: to find probability of proton being in on coordinate $(43,21,65)$ and electron being in coordinates (43,2,12) at time 5 is $P(5, 43,21,65,43,2,12)=|\psi_{example}(5, 43,21,65,43,2,12)|^2$

operator applied to a wavefunction equals to wavefunction times the value that the operator is measuring. $\hat{O}(\psi)(arguments\ of\ psi)=\psi(arguments\ of\ psi)\ \lambda(arguments\ of\ psi)$ .
for example: if in the example-system I apply momentum operator to get projection of momentum of electron to the first coordinate direction, then $\hat{O}_{get\ x-momentum\ of\ electron}(\psi_{example})=-i\ h/(2pi)\ \frac{\partial \psi_{example}}{\partial x_{1.1}}=p_{1.1}\ \psi=m_{electron}\ v_{1.1}\ \psi$

this equation can be simplified to $\hat{O}(\psi)(arguments\ of\ psi)/\psi(arguments\ of\ psi)=\lambda(arguments\ of\ psi)$ where lambda is the quantity, that the operator is measuring.
The equation about the example-system can be simplified to $p_{1.1}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})=-\frac{i\ h}{2pi\ \psi_{example}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})}\ \frac{\partial \psi_{example}}{\partial x_{1.1}}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})$ .

If $\psi$ is eigenfunction of operator $\hat{O}$ then $\lambda$ is the eigenvalue and does not depend of arguments of $\psi$ and the quantity that the $\hat{O}$ measures is same in every time and every place in the system.
For example if projection of momentum of the electron to the 1. coordinate is known and does not depend of place and time. Then $p_{1.1}$ does not depend of $x_{0.1}$ , $x_{0.2}$ , $x_{0.3}$ , $x_{1.1}$ , $x_{1.2}$ nor $x_{1.3}$. It can be so only if wavefunction $\psi_{example}$can be written in the form $\psi_{example}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})=F(t,x_{0.1},x_{0.2},x_{0.3},x_{1.2},x_{1.3})\ e^{C*x_{1.1}}$ where F is an arbitary function and C is an arbitary compleksnumber. In this case the electron is equally likely to have any first-coordinate value.

If $\psi$ is not eigenfunction of $\hat{O}$ then the value, that the operator $\hat{O}$ measures, does depend of generalized coordinates of the system.
for example in example-system momentum of the electron
$p_{1.1}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})=-\frac{i\ h}{2pi\ \psi_{example}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})}\ \frac{\partial \psi_{example}}{\partial x_{1.1}}(t,x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})$ does depend of time and coordinates of proton and electron.

 

[anuttarasammyak]

to the first coordinate direction, then O^get x−momentum of electron(ψexample)=−i h/(2pi) ∂ψexample∂x1.1=p1.1 ψ=melectron v1.1 ψ

this equation can be simplified to

This holds only when $\psi$ is an eigenfunction of momentum of electron with eigenvalue p1.1.
Usually $\psi$ is superposition of various momentum eigenstates.

 

[olgerm]

View attachment 320999
This holds only when $\psi$ is an eigenfunction of momentum of electron with eigenvalue p1.1.
Usually $\psi$ is superposition of various momentum eigenstates.

How to find information about projection of momentum of electron to the first coordinate-axis if $\psi$ is not eigenfunction of momentum of electron with eigenvalue p1.1?
can $\psi$ be eigenfunction of momentum of electron with eigenvalue p1.1 if the electron is not equally likely to have any 1.-coordinate value (because Heisenberg uncertainty principle)?

 

[DrClaude]

to find probility of proton being in on coordinate $(43,21,65)$ and electron being in coorintes (43,2,12) at time 5 is $P(5, 43,21,65,43,2,12)=|\psi_{example}(5, 43,21,65,43,2,12)|^2$

Probability densities are made to be integrated. The probability of finding a particle at a particular set of coordinates should always be 0 (as there are an infinity of possible coordinates).

Similarly, I can't make sense of an operator acting on a wave function at a single coordinate. Operators operate on the entire state of the system, in this case the full wave function.

 

[anuttarasammyak]

How to find information about projection of momentum of electron to the first coordinate-axis if ψ is not eigenfunction of momentum of electron with eigenvalue p1.1?

In order to get information of momentum from coordinate wave function, we usually calculate
$$<p^n>=\int \psi^*(x) \hat{p}^n \psi(x) dx$$
where momentum operator
$$\hat{p}=\frac{\hbar }{i }\frac{\partial}{\partial x}$$
and LHS $<p^n>$ is expectation value of $p^n$. If you transform coordinate wave function to momentum wave function $\phi(p)$ by Fourier transform, the integrand becomes just multiplication
$$<p^n>=\int \phi^*(p) p^n \phi(p) dp$$
which might be convenient for your understanding where $|\phi(p)|^2$ is probability density of p.
$\phi(p)$ gives superposition weight of various momentum eigenstates for our state..

 

[olgerm]

Probability densities are made to be integrated. The probability of finding a particle at a particular set of coordinates should always be 0 (as there are an infinity of possible coordinates).

Yes, sure. I meant "to find probability-density of proton being on coordinate" not "to find probility of proton being in on coordinate".

 

[olgerm]

In order to get information of momentum from coordinate wave function, we usually calculate
$$<p^n>=\int \psi^*(x) \hat{p}^n \psi(x) dx$$

is the integral in this equation taken over all arguments of the wavefunction (including time)? So for the example I wrote there would be 7 integrals from $-\infty$ to $\infty$?

 

[andresB]

is the intgral in this equation taken over all arguments of the wavefunction (including time)? So for the example I wrote there would be 7 integrals form $-\infty$ to $\infty$?

The integral is over the position coordinates, the time argument remains untouched. Therefore, the expectation value is a function of time.

 

[PeroK]

is the intgral in this equation taken over all arguments of the wavefunction (including time)? So for the example I wrote there would be 7 integrals form $-\infty$ to $\infty$?

An integral over time would diverge.

 

[olgerm]

If you transform coordinate wave function to momentum wave function $\phi(p)$ by Fourier transform, the integrand becomes just multiplication
$$<p^n>=\int \phi^*(p) p^n \phi(p) dp$$
which might be convenient for your understanding where $|\phi(p)|^2$ is probability density of p.
$\phi(p)$ gives superposition weight of various momentum eigenstates for our state..

May it be that in some systems expectation value of momentum of a particle depends of coordinates of the particle? If so how to get probability density that has both momentums and coordinates as arguments?

 

[anuttarasammyak]

May it be that in some systems expectation value of momentum of a particle depends of coordinates of the particle? If so how to get probability density that has both momentums and coordinates as arguments?

We get <p>, <f(p)>, <x>, <g(x)>, etc. from coordinate wavefunction $\psi(x)$ which is function of x or momentum wavefuncton $\phi(p)$ which is function of p and often noted as $\psi(p)$ also. The two wavefuncitons are equivalent. In QM we cannot appoint the system to have such coordinate and such momentum also.

 

[PeroK]

May it be that in some systems expectation value of momentum of a particle depends of coordinates of the particle? If so how to get probability density that has both momentums and coordinates as arguments?

The wavefunction represents the state of the particle, which determines the distribution of position and momentum measurement outcomes. It can be expressed either as a function of position, or a function of momentum.

 

[vanhees71]

May it be that in some systems expectation value of momentum of a particle depends of coordinates of the particle? If so how to get probability density that has both momentums and coordinates as arguments?

The expectation value of any observable cannot depend on coordinates of the particle, because expectation values are calculated by integrating over all space (when expressed in the position representation). In fact it depends only on the state, not on any representation. For a pure state you can choose an arbitrary ket, $|\psi \rangle$ representing the state $\hat{\rho}=|\psi \rangle \langle \psi|$, and you get for the expectation value of any observable $A$, represented by the corresponding self-adjoint operator $\hat{A}$,
$$\langle A \rangle = \langle \psi|hat{A}|\psi \rangle.$$
In the position representation this reads
$$\langle A \rangle = \int_{\mathbb{R}^3} \mathrm{d} x \psi^*(t,x) \hat{A} \psi(t,x),$$
and that's indeed a function of time but, of course, not of any position $x$.

 

[olgerm]

The expectation value of any observable cannot depend on coordinates of the particle

Imagine following system: a particle is in a potential-well. $(E_{potential}(x)=\infty\ if\ x<0) \land (E_{potential}(x)=0\ if\ 0<x\ \land\ x<1)\land (E_{potential}(x)=1\ if\ 1<x\ \land\ x<2)\land (E_{potential}(x)=\infty\ if\ 2<x)$ Potential in one side of the potential-well is bigger than potential in the other side. Energy of the particle is bigger than the potential(energy of the particle) in the 2. side(1<x and x<2) of the potential-well. The particle has some likelihood of being in either side of the well.
In non-quantum-physics potential energy of the particle would be smaller if it were in 1. side(0<x and x<1) of the potential-well. Kinetic energy of the particle would be bigger if it were in the 1. side of the potential-well.
Would not the momentum of the particle also be bigger if it were in 1. side of the well? Would not the momentum of the particle also be bigger in the 1. side of the well?

 

[anuttarasammyak]

Imagine following system: a particle is in a potential-well.

As I said in QM "where" and "how much momentum" are questions to be posed but "how much momentum where the particle is" is a wrong question.

Mometum of a particle in energy eigenstate of a infinite well system is a popular exercise. See e.g., https://phys.libretexts.org/Courses/University_of_California_Davis/UCD:_Physics_9HE_-_Modern_Physics/03:_One-Dimensional_Potentials/3.2:_Infinite_Square_Well

 

[vanhees71]

A particle in an infinite well has no momentum observable. Of course you can use a more realistic potential to trap your particle like harmonic-oscillator potential, where the particle has a well-defined momentum observable to demonstrate the Heisenberg uncertainty relation for position and momentum for harmonic-oscillator energy eigenstates.

 

[olgerm]

Sorry, I have not been studied for long time now. Maybe my thought was very stupid. I thought like that: from wavefunction we can get probability-density of were particle could be. And now for every point of space we could calculate kinetic energy of particle knowing the potential-field, using classical physics and treating the particle like a point-mass-charge.
For example for system with constant potential and 1 particle: if from wavefunction we know that the particle has 70% probaility of being in area $x<0$, where its potential energy would be $2*10^-25*J$ and 30% of probabilty of being in area $0<x$, where its potential energy would be $6*10^-25*J$ and that the total energy in system is $10^-24*J$ then we could say that the particle is
70% likely to have x-coordinate $x<0$ and kinetic energy $8*10^-25*J$ and
30% likely to have x-coordinate $0<x$ and kinetic energy $4*10^-25*J$.

 

[PeroK]

Sorry, I have not been studied for long time now. Maybe my thought was very stupid. I thought like that: from wavefunction we can get probability-density of were particle could be. And now for every point of space we could calculate kinetic energy of particle knowing the potential-field, using classical physics and treating the particle like a point-mass-charge.
For example for system with constant potential and 1 particle: if from wavefunction we know that the particle has 70% probaility of being in area $x<0$, where its potential energy would be $2*10^-25*J$ and 30% of probabilty of being in area $0<x$, where its potential energy would be $6*10^-25*J$ and that the total energy in system is $10^-24*J$ then we could say that the particle is
70% likely to have x-coordinate $x<0$ and kinetic energy $8*10^-25*J$ and
30% likely to have x-coordinate $0<x$ and kinetic energy $4*10^-25*J$.

That's a classical model of a particle. It is definitely somewhere with a given probability distribution; and, at every point in space, there is a definite kinetic energy (derivable from conservation of energy and the potential at that point).

That might be a model for a classical pendulum with uncertain position. But, the quantum harmonic oscillator, for example, has an entirely different theoretical model. It's not simply a classical oscillator where the position is probabilistically distributed.

In general, all your posts on QM involve trying to force QM into a classical framework. That won't work.

 

[olgerm]

And it also would not be possible to find expectationvalue of A in a specific region of space like that:
$$<A>=\int_{region}(\psi^*(x) \hat{A} \psi(x) dx) / \int_{region}(\psi^*(x) \psi(x) dx)$$
?
Integral is not over all space , but only over a specific region of space here.

 

[PeroK]

And it also would not be possible to find expectationvalue of A in a specific region of space like that:
$$<A>=\int_{region}(\psi^*(x) \hat{A} \psi(x) dx) / \int_{region}(\psi^*(x) \psi(x) dx)$$
?
Integral is not over all space , but only over a specific region of space here.

You have to think about how physically meaningful (or not) that restricted expectation value is.

 

[olgerm]

You have to think about how physically meaningful (or not) that restricted expectation value is.

I do not really know. That is why I asked it. I guessed if someone measured electron this given region, then it he would have that expectation value.

70% likely to have x-coordinate $x<0$ and kinetic energy $8*10^-25*J$ and
30% likely to have x-coordinate $0<x$ and kinetic energy $4*10^-25*J$.

That might be a model for a classical pendulum with uncertain position. But, the quantum harmonic oscillator, for example, has an entirely different theoretical model.

So if some one measured the energy of the electron in area x<0 he would have the same expectation value as he measured these in region 0<x?

 

[olgerm]

$|\phi(p)|^2$ is probability density of p.

It is easier to understand it as functions(operators that take wavefunction as argument) and return new functions, which module square is probabity-density of the observable value. Is there a function(like fourier transform is to get momentum-wavefunction form position-wavefunction) that can produce a new (wave)function, which module is probabilty-density of corresponding observable value? For example how to get a function $\psi_{angular}$, which module square would be probability-density of angularmomentum ($|\psi_{angular}|^2=p(t,L_{1},L_{2},L_{3})$ for 1 particle system)?

 

[PeroK]

I do not really know. That is why I asked it. I guessed if someone measured electron this given region, then it he would have that expectation value.

So if some one measured the energy of the electron in area x<0 he would have the same expectation value as he measured these in region 0<x?

I suggest you get a QM textbook and study it. Making up your own hybrid classical-QM questions will get you nowhere.

 

[olgerm]

I suggest you get a QM textbook and study it. Making up your own hybrid classical-QM questions will get you nowhere.

I try to, but could you just directly answer my question? Getting direct and clear answers to such basic questions makes learning much faster. Which book do you recommend?

 

[PeroK]

I try to, but could you just directly answer my question?

QM isn't a branch of classical mechanics. You can't ask questions based on classical ideas and expect them to have clear answers.

Getting direct and clear answers to such basic questions makes learning much faster. Which book do you recommend?

There are lots of recommendations on here. Griffiths, for example, starts with wave mechanics. That's the only introductory book I have.

I also have Sakurai, which starts with quantum spin and the Dirac formalism. That's slightly mpore advanced, but it may be a good option for you.

There are also these very insightful notes and may give you a deeper understanding of QM from the beginning:

http://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

 

[vanhees71]

It is easier to understand it as functions(operators that take wavefunction as argument) and return new functions, which module square is probabity-density of the observable value. Is there a function(like fourier transform is to get momentum-wavefunction form position-wavefunction) that can produce a new (wave)function, which module is probabilty-density of corresponding observable value? For example how to get a function $\psi_{angular}$, which module square would be probability-density of angularmomentum ($|\psi_{angular}|^2=p(t,L_{1},L_{2},L_{3})$ for 1 particle system)?

That becomes very clear in Dirac's formulation of QT. It describes QT on an abstract Hilbert space:

(a) A (pure) state of a quantum system is described by a ray in Hilbert space, i.e., a vector $|\psi \rangle \neq 0$ modulo an arbitrary non-zero complex number.

(b) An observable is represented by a self-adjoint operator acting on Hilbert-space vectors. The possible outcomes of a measurement of an observable are the eigenvalues of these operators.

Then math tells you that there are complete sets of (generalized) orthonormal eigenvectors (CONS) of self-adjoint operators, and that there are CONSs of common eigenvectors of two such operators if and only if these two operators commute. This enables the definition

(c) A complete set of compatible observables is given by a corresponding set of commuting self-adjoint operators whose common eigenvectors are unique (up to a factor), i.e., if the common eigenspaces are all one-dimensional.

With this we can state the meaning of the state vectors:

(d) If the self-adjoint $\hat{A}_1,\ldots,\hat{A}_n$ are representing a complete set of compatible observables, and $|a_1,\ldots,a_n \rangle$ are a complete set of orthonormalized common eigenvectors, then the probability for finding the possible values $(a_1,\ldots,a_n)$ when measuring these observables is given by
$$P(a_1,\ldots,a_n)=\frac{1}{\langle \psi|\psi \rangle} |\langle a_1,\ldots,a_n|\psi \rangle|^2,$$
where $|\psi \rangle$ represents the state the measured system is prepared in when measured.

In non-relativistic quantum theory of a particle with spin 0 a complete set of compatible observables are, e.g., the three components of the position vector of the particle, $\vec{x}=(x_1,x_2,x_3)$. From the commutator algebra between position and momentum, which defines the entire algebra of observables in this case,
$$[\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \hat{1}$$
one deduces that both the $\hat{x}_k$ and $\hat{p}_j$ have all real numbers as "eigenvalues", and the corresponding "eigenvectors" are in fact not true Hilbert-space vectors but distributions (the math is a bit subtle; a good introduction to it is given in the textbook Ballentine, Quantum Mechanics), i.e., they are "orthonoramlizable" only in the sense of "Dirac distributions",
$$\langle \vec{p}_1|\vec{p}_2 \rangle=\delta^{(3)}(\vec{p}_1-\vec{p}_2), \quad \langle \vec{x}_1|\vec{x}_2 \rangle=\delta^{(3)}(\vec{x}_1-\vec{x}_2).$$
In the following we also assume that the state ket $|\psi \rangle$, representing the state the particle is prepared in, is normalized, $\langle \psi|\psi \rangle=1$. Then the position representation of the state, is the wave function in Schrödingers wave-mechanics formulation,
$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle,$$
and $P(\vec{x})=|\psi(\vec{x})|^2$ is the probability density of position.

It's also very clear that the probability density of momentum is given by
$$\tilde{\psi}(\vec{p})=\langle \vec{p}|\psi \rangle.$$
With the completeness of the momentum "eigenstates" you immediately get
$$\tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 x \langle \vec{p} | \vec{x} \rangle \langle \vec{x}|\psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 x \langle \langle \vec{p} | \vec{x} \rangle \psi(\vec{x}).$$
Now, again using the above commutator relations between position and momentum components you can show that
$$\langle \vec{p}|\vec{x} \rangle=\frac{1}{(2 \pi \hbar)^{3/2}} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}/\hbar),$$
and this explains why the "position-wave function" is the Fourier transform of the "momentum-wave function" and vice versa:
$$\tilde{\psi}(\vec{p}) =\int_{\mathbb{R}^3} \mathrm{d}^3 x \frac{1}{(2 \pi)^{3/2}} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}/\hbar) \psi(\vec{x})$$
and in the same way the other way around
$$\psi(\vec{x}) =\int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{(2 \pi)^{3/2}} \exp(+\mathrm{i} \vec{p} \cdot \vec{x}/\hbar) \tilde{\psi}(\vec{p}),$$
where I used
$$\langle \vec{x}|\vec{p} \rangle=\langle \vec{p}|\vec{x} \rangle^*.$$
You can of course choose any other set of compatible observables, e.g., $|\vec{x}|=r$ and $\vec{L}^2$ and $\vec{L}_z$, where $\vec{L}=\vec{x} \times \vec{p}$ is the orbital angular momentum of the particle (wrt. the origin of the coordinate system). Note that since
$$[\hat{L}_j,\hat{L}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{L}_l$$
the three angular-momentum components are NOT compatible observables, and thus there's no common CONS of eigenvectors of all three angular-momentum components, but you can show that indeed
$$[\hat{r},\hat{L}_j]=0, \quad [\hat{\vec{L}}^2,\hat{L}_j]=0,$$
and thus $r$, $\hat{\vec{L}}^2$ as well as $\hat{L}_3$ are a compatible set of observables which define a common CONS of eigenvectors. Thus you can define the corresponding "wave functions"
$$\psi(r,\ell,m)=\langle r,\ell,m|\psi \rangle,$$
where
$$\hat{r} |r,\ell,m \rangle, \quad \hat{\vec{L}}^2 |r,\ell,m \rangle=\hbar^2 \ell(\ell+1) |r,\ell,m \rangle, \quad \hat{L}_z |r,\ell,m \rangle=\hbar m |r,\ell,m \rangle,$$
where the eigenvalues are $r \in \mathbb{R}_{\geq 0}$, $\ell \in \mathbb{N}_0=\{0,1,2,\ldots \}$, and for this given $\ell$ the $m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}$.

You can NOT have wave functions of the type $\psi(\vec{L})$ (NONSENSE!), because the $\hat{\vec{L}}$ are not commuting and there's thus no CONS of common eigenvectors of this set of observables!

 

[PeroK]

Who needs a textbook when you have @vanhees71 !

 

[vanhees71]

Well, this is not more than an appetizer to read a textbook to understand it. It's about 3/4 of a standard QM lecture in a short forum posting.

My recommendation as the introductory QM textbook is

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics,
Addison Wesley (1993).

The more recent edition co-edited by Napolitano is also not distorting the content of this edition as far as I remember, but it contains a superfluous addendum of a socalled "relativistic quantum mechanics", and I've a strict principle to never recommend any textbook with such a misleading content. The only consistent relativistic QT is relativistic QFT, and "relativistic QM" is only very confusing the subject, which is anyway pretty subtle!

 

[anuttarasammyak]

For example for system with constant potential and 1 particle: if from wavefunction we know that the particle has 70% probaility of being in area x<0, where its potential energy would be 2∗10−25∗J and 30% of probabilty of being in area 0<x, where its potential energy would be 6∗10−25∗J and that the total energy in system is 10−24∗J then we could say that the particle is
70% likely to have x-coordinate x<0 and kinetic energy 8∗10−25∗J and
30% likely to have x-coordinate 0<x and kinetic energy 4∗10−25∗J.

@olgerm Have you solved the exercise of finite well potential e.g., Wiki https://en.wikipedia.org/wiki/Finite_potential_well ? It would give you a good insight.
In solving Shrodeinger equation, two parts in the well and in the wall are connected.
In the well : $e^{ikx}$ wave type
In the wall : $e^{-\kappa x}$ exponential decay type,i.e. permutation into the wall or tunneling effect when wavfunction go through to another free region beyond if any.

With momentum eigenfuction $e^{ikx}, k=p/\hbar, -\infty<x<+\infty$ in mind, we are tempted to think "In the well the particle has classical to and fro momentum and in the wall it has imaginary momentum which generates negative kinetic energy".　What do you think of this ?

 

[vanhees71]

It's the wrong way to think about QT. You calculate the energy eigenvectors and eigenvalues, i.e., you determine in which states you particle has a definite energy. Since $[\hat{H},\hat{p}]\neq 0$, usually these are not eigenvectors of $\hat{p}$, i.e., the momentum has no definite value, but you can only predict the probabilities which result you get when measuring the momentum of the particle.

 

[olgerm]

My recommendation as the introductory QM textbook is

J. J. Sakurai and S. Tuan, Modern Quantum Mechanics,
Addison Wesley (1993).

I took "David J. Griffith, Introduction to Quantum Mechanics". Also tried to read "J. J. Sakurai and S. Tuan, Modern Quantum Mechanics", but "David J. Griffith, Introduction to Quantum Mechanics" seems more easily understandable.

 

[kered rettop]

Who needs a textbook when you have @vanhees71 !

True, but sometimes having @vanhees71, makes you realize you should have read your textbook.