Mixing
Why position and momentum operators are both continuous spectrum while angular momentum is discrete? [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
This question already has an answer here:
Why does spin have a discrete spectrum?
3 answers
We know that position $hatr$ and momentum $hatp$ are both continuous spectrum operators, i.e.
$$hatr|r'rangle=r'|r'rangle, quad hatp|p'rangle=p'|p'rangle.$$
But the angular operator $hatL=hatrtimeshatp$ is not:
$$hatL^2|lrangle=hbar^2 l(l+1)|lrangle.$$
Could anyone give some explanation?
quantum-mechanics angular-momentum operators discrete
edited 8 mins ago
Qmechanic♦
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
marked as duplicate by Aaron Stevens, Qmechanic♦ quantum-mechanics
Users with the  quantum-mechanics badge can single-handedly close quantum-mechanics questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
3 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
4
down vote
favorite
This question already has an answer here:
Why does spin have a discrete spectrum?
3 answers
We know that position $hatr$ and momentum $hatp$ are both continuous spectrum operators, i.e.
$$hatr|r'rangle=r'|r'rangle, quad hatp|p'rangle=p'|p'rangle.$$
But the angular operator $hatL=hatrtimeshatp$ is not:
$$hatL^2|lrangle=hbar^2 l(l+1)|lrangle.$$
Could anyone give some explanation?
quantum-mechanics angular-momentum operators discrete
edited 8 mins ago
Qmechanic♦
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
marked as duplicate by Aaron Stevens, Qmechanic♦ quantum-mechanics
Users with the  quantum-mechanics badge can single-handedly close quantum-mechanics questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
3 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
Why does spin have a discrete spectrum?
3 answers
We know that position $hatr$ and momentum $hatp$ are both continuous spectrum operators, i.e.
$$hatr|r'rangle=r'|r'rangle, quad hatp|p'rangle=p'|p'rangle.$$
But the angular operator $hatL=hatrtimeshatp$ is not:
$$hatL^2|lrangle=hbar^2 l(l+1)|lrangle.$$
Could anyone give some explanation?
quantum-mechanics angular-momentum operators discrete
edited 8 mins ago
Qmechanic♦
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
This question already has an answer here:
Why does spin have a discrete spectrum?
3 answers
We know that position $hatr$ and momentum $hatp$ are both continuous spectrum operators, i.e.
$$hatr|r'rangle=r'|r'rangle, quad hatp|p'rangle=p'|p'rangle.$$
But the angular operator $hatL=hatrtimeshatp$ is not:
$$hatL^2|lrangle=hbar^2 l(l+1)|lrangle.$$
Could anyone give some explanation?
This question already has an answer here:
Why does spin have a discrete spectrum?
3 answers
quantum-mechanics angular-momentum operators discrete
quantum-mechanics angular-momentum operators discrete
edited 8 mins ago
Qmechanic♦
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
edited 8 mins ago
Qmechanic♦
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
edited 8 mins ago
Qmechanic♦
97.9k121661059
edited 8 mins ago
Qmechanic♦
97.9k121661059
edited 8 mins ago
Qmechanic♦
97.9k121661059
97.9k121661059
asked 26 mins ago
Stone-Zeng
22417
asked 26 mins ago
Stone-Zeng
22417
asked 26 mins ago
Stone-Zeng
22417
22417
marked as duplicate by Aaron Stevens, Qmechanic♦ quantum-mechanics
Users with the  quantum-mechanics badge can single-handedly close quantum-mechanics questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
3 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Aaron Stevens, Qmechanic♦ quantum-mechanics
Users with the  quantum-mechanics badge can single-handedly close quantum-mechanics questions as duplicates and reopen them as needed.
StackExchange.ready(function()
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();
);
);
);
3 mins ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago
add a comment |Â
4
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago
4
4
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
answered 13 mins ago
levitopher
3,94211435
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
add a comment |Â
up vote
0
down vote
In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2pi$, therefore its eigenvalues are also discrete.
answered 8 mins ago
Gradient137
246
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
answered 13 mins ago
levitopher
3,94211435
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
add a comment |Â
up vote
1
down vote
One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
answered 13 mins ago
levitopher
3,94211435
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
answered 13 mins ago
levitopher
3,94211435
One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
answered 13 mins ago
levitopher
3,94211435
answered 13 mins ago
levitopher
3,94211435
answered 13 mins ago
levitopher
3,94211435
answered 13 mins ago
levitopher
3,94211435
3,94211435
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
add a comment |Â
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
But even in bound states position is continuous.
– Aaron Stevens
6 mins ago
add a comment |Â
up vote
0
down vote
In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2pi$, therefore its eigenvalues are also discrete.
answered 8 mins ago
Gradient137
246
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
add a comment |Â
up vote
0
down vote
In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2pi$, therefore its eigenvalues are also discrete.
answered 8 mins ago
Gradient137
246
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2pi$, therefore its eigenvalues are also discrete.
answered 8 mins ago
Gradient137
246
In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2pi$, therefore its eigenvalues are also discrete.
answered 8 mins ago
Gradient137
246
answered 8 mins ago
Gradient137
246
answered 8 mins ago
Gradient137
246
answered 8 mins ago
Gradient137
246
246
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
add a comment |Â
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
But even in bound states position is continuous.
– Aaron Stevens
1 min ago
add a comment |Â
4
Perhaps it is related to the boundary conditions?
– K_inverse
18 mins ago
Are you asking for a mathematical reason from QM, or are you asking why the universe works this way?
– Aaron Stevens
16 mins ago
Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein.
– Qmechanic♦
8 mins ago
Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous.
– Aaron Stevens
2 mins ago