A very basic question about bra-kets

  • I
  • Thread starter olgerm
  • Start date
In summary: But in conventional quantum theory, the wave function is defined as a function of both space and time, and the 'bra' state-vector is defined on a domain of space and time (spanned by the 'ket's.)
  • #1
olgerm
Gold Member
533
35
Do I understand correctly that of a system bra in basis of coordinates is a vector of which components are in bijective relation with possible values of the wavefunction that describes the same system? And that the value of every components of bra is the value that the vawefunction would return if it had these values as arguments?

For example for a system that has 1 electron and 1 proton:
wavefunction would be:

##\psi_{example}(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})##
  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.
bra, for this system would have ##|Re|^7## components, where ##|Re|## is cardinality of set of all real numbers.
for example the components of bra that corresponds to values arguments of the wavefunction ##(t=9, x_{0.1}=23,x_{0.2}=43,x_{0.3}=83,x_{1.1}=21,x_{1.2}=87,x_{1.3}=93)## has value ##\psi_{example}(9, 23,43,83,21,87,93)##?
 
Physics news on Phys.org
  • #2
Bras and kets are part of an abstract formalism of QM that reduces to wave mechanics in the position or momentum basis.

If you are learning about the abstract formalism, then I can make no sense of a construction like:
olgerm said:
##\psi_{example}(9, 23,43,83,21,87,93)##
 
  • Like
Likes topsquark, vanhees71 and DrClaude
  • #3
@olgerm Each componets has its bra vector so
[tex]|x_{e1}>,|x_{e2}>,|x_{e3}>,|x_{p1}>,|x_{p2}>,|x_{p3}>[/tex]
Their product
[tex]|x_{e1}>|x_{e2}>|x_{e3}>|x_{p1}>|x_{p2}>|x_{p3}>[/tex]
forms bra vector of the system.
Wave function corresponds to the state vector
[tex]\int dx_{e1} \int dx_{e2}\int dx_{e3}\int dx_{p1}\int dx_{p2}\int dx_{p3}\ \psi |x_{e1}>|x_ {e2}>|x_{e3}>|x_{p1}>|x_{p2}>|x_{p3}>[/tex]
where
[tex]\psi=\psi(t, x_{e1},x_{e2},x_{e3},x_{p1},x_{p2},x_{p3})[/tex]

Q. Get probality to observe the electrion in volume ##\triangle V_e## and the positron ##\triangle V_p##.

Ans.
[tex]\int_{\triangle V_e}dV_e\int_{\triangle V_p}dV_p \int_{\triangle V_e}dV'_e\int_{\triangle V_p}dV'_p [/tex][tex]\ <x' _{e1}|<x'_{e2}|<x'_{e3}|<x'_{p1}|<x'_{p2}|<x'_{p3}|\ \psi'^ * \psi |x_{e1}>|x_{e2}>|x_{e3}>|x_{p1}>|x_{p2}>|x_{p3}>[/tex]
[tex]=\int_{\triangle V_e}dV_e\int_{\triangle V_p}dV_p\ |\psi|^2[/tex]
where
[tex]dV_e=dx_{e1}dx_{e2}dx_{e3}[/tex]
[tex]dV'_e=dx'_{e1}dx'_{e2}dx'_{e3}[/tex]
[tex]dV_p=dx_{p1}dx_{p2}dx_{p3}[/tex]
[tex]dV'_p=dx'_{p1}dx'_{p2}dx'_{p3}[/tex]
and
[tex]\psi=\psi(t, x_{e1},x_{e2},x_{e3},x_{p1},x_{p2},x_{p3})[/tex]
[tex]\psi'=\psi(t, x'_{e1},x'_{e2},x'_{e3},x'_{p1},x'_{p2},x'_ {p3})[/tex]

olgerm said:
for example the components of bra that corresponds to values arguments of the wavefunction (t=9,x0.1=23,x0.2=43,x0.3=83,x1.1=21,x1.2=87,x1.3=93) has value ψexample(9,23,43,83,21,87,93)?
Thus waqvefunction of the state which has definite coordinates ##(E_1,E_2, E_3, P_1,P_2,P_3) ## is
[tex]\psi=\delta(x_{e1}-E_1) \delta(x_{e2}-E_2) \delta(x_{e3}-E_3)\delta(x_{p1}-P_1)\delta(x_ {p2}-P_2)\delta(x_{p3}-P_3)[/tex]

As the ket vector
[tex]\int dx_{e1} \int dx_{e2}\int dx_{e3}\int dx_{p1}\int dx_{p2}\int dx_{p3}\ |x_{e1}>|x_{e2 }>|x_{e3}>|x_{p1}>|x_{p2}>|x_{p3}>[/tex][tex]\ \delta(x_{e1}-E_1) \delta(x_{e2 }-E_2) \delta(x_{e3}-E_3)\delta(x_{p1}-P_1)\delta(x_{p2}-P_2)\delta(x_{p3}-P_3) [/tex]
simply
[tex]=|E_1>|E_2>|E_3>|P_1>|P_2>|P_3>[/tex]

[tex]\int_{any\ region \ including(E_1,E_2, E_3, P_1,P_2,P_3)} dV_e dV_p |\psi|^2=\delta(0)^6=\infty[/tex]
[tex]\int_{any\ region \ excluding (E_1,E_2, E_3, P_1,P_2,P_3)} dV_e dV_p |\psi|^2=0[/tex]

Unfortunately the coordinate eigenstate as well as other continuous observables' cannot be normalized to one.
 
Last edited:
  • #4
olgerm said:
Do I understand correctly that of a system bra in basis of coordinates is a vector of which components are in bijective relation with possible values of the wavefunction that describes the same system? And that the value of every components of bra is the value that the vawefunction would return if it had these values as arguments?

For example for a system that has 1 electron and 1 proton:
wavefunction would be:

##\psi_{example}(t, x_{0.1},x_{0.2},x_{0.3},x_{1.1},x_{1.2},x_{1.3})##
  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.
bra, for this system would have ##|Re|^7## components, where ##|Re|## is cardinality of set of all real numbers.
for example the components of bra that corresponds to values arguments of the wavefunction ##(t=9, x_{0.1}=23,x_{0.2}=43,x_{0.3}=83,x_{1.1}=21,x_{1.2}=87,x_{1.3}=93)## has value ##\psi_{example}(9, 23,43,83,21,87,93)##?
In conventional quantum mechanics, a wave function is a time dependent function over a set of (possibly time dependent) coordinates (e.g. position, and in this case, spin.) In your example, \emph{ignoring spin} the Hilbert space is spanned by 'bras' of the form ##|\vec x_e, \vec x_p\rangle## , or \emph{with spin, ignoring quark degrees of freedom}, ##|\vec x_e, \sigma_e, \vec x_p, \sigma_p\rangle##. In practice, the inner product of this basis (with spin) is defined by linear extension of (letting ##\vec Q## denote continuous variables ##(\vec x_e, \vec x_p)## and ##\vec \sigma## the discrete spin variables ##(\sigma_e,\sigma_p)##) ##\langle \vec Q_1, \vec\sigma_1|\vec Q_2,\vec\sigma_2\rangle = \delta^6(\vec Q_1-\vec Q_2)\delta_{\vec\sigma_1,\vec\sigma_2}##, where ##\delta_{\vec\sigma_1,\vec\sigma_2}## is the Kronecker delta over ##(\sigma_e,\sigma_p)##. Hence, (accounting for spin) there are (informally speaking) "##2^2|Re|^6##", (where ##|Re|=|\mathbb R|##) linearly independent basis states. Because an electron is distinct from a proton (by mass or charge), there is no restriction on the coordinates of either (i.e. ##\vec x_e## and ##\vec x_p## both cover ##\mathbb R^3##.

In a different formulation of quantum mechanics, you could construct a theory involving 'wave functions' for 'events' distributed over space as well as time, but that would be a complete departure from quantum theory and require a completely new framework (and special attention dedicated to how the form of the wave function for an 'event' would be measured/constrained.)
 
  • #5
In standard quantum mechanics a wave function is a function of time and configuration-space variables, which are independent from each other. The configuration-space variables are not functions of time.

The relation to the representation-free formulation is as follows. In the Heisenberg picture of time evolution the state ket evolves with time due to the full Hamiltonian of the system,
$$\mathrm{i} \hbar \partial_t |\psi,t \rangle=\hat{H} |\psi,t \rangle.$$
The wave function are then the components of ##|\psi,t \rangle## wrt. a complete orthornormal system of (generalized) common eigenvectors of a minimal complete set of observable-operators ##\hat{A}_k##,
$$\psi(t,a_1,\ldots,a_n)=\langle a_1,\ldots,a_n|\psi,t \rangle.$$
E.g., for a particle you can choose the position-vector components and the spin component ##\sigma_z## as a complete set of compatible observables. Then you get the wave function in the position representation,
$$\psi(t,\vec{x},\sigma_3)=\langle \vec{x},\sigma_3|\psi,t \rangle.$$
 
  • #6
olgerm said:
for this system would have ##|Re|^7## components, where ##|Re|## is cardinality of set of all real numbers.
It's completely irrelevant here, but I want to note that ##\mathbb{R}^n## has the same cardinality as ##\mathbb{R}##.
 
  • Like
Likes vanhees71
  • #7
Demystifier said:
It's completely irrelevant here, but I want to note that ##\mathbb{R}^n## has the same cardinality as ##\mathbb{R}##.
Yeah, cardinality is not the appropriate concept here. You need topology, in some form, to make more meaningful statement about dimensions of degrees of freedom.
 
  • Like
Likes kered rettop, vanhees71 and Demystifier
  • #8
vanhees71 said:
In standard quantum mechanics a wave function is a function of time and configuration-space variables, which are independent from each other. The configuration-space variables are not functions of time.
Is it possible to have a wave function in a non-inertial (e.g. co-rotating) reference frame?
 
  • #9
Couchyam said:
Is it possible to have a wave function in a non-inertial (e.g. co-rotating) reference frame?
The quantum state can be transformed to any frame, but there's rarely call to do it pragmatically because it would make calculations difficult.
 
  • Like
Likes vanhees71
  • #10
LittleSchwinger said:
The quantum state can be transformed to any frame, but there's rarely call to do it pragmatically because it would make calculations difficult.
I thought the canonical example was from basic NMR physics: the effect of a resonant pulse is often treated (for simplicity) in the rest frame of a spin undergoing Larmor precession, and that's supposed to get students thinking about how non-inertial reference frames are treated in QM.
 
  • #11
Couchyam said:
I thought the canonical example was from basic NMR physics: the effect of a resonant pulse is often treated (for simplicity) in the rest frame of a spin undergoing Larmor precession, and that's supposed to get students thinking about how non-inertial reference frames are treated in QM.
Usually that's still treated in an inertial frame. Of course you don't have to, it's just easier.
 

Similar threads

Replies
31
Views
3K
Replies
19
Views
3K
Replies
9
Views
2K
Replies
17
Views
1K
Replies
1
Views
3K
Replies
8
Views
475
Replies
24
Views
2K
Replies
16
Views
2K
Back
Top