I am really sorry for troubling you guys with this simple problem but it has bugging me for a while and I know I got it wrong in some really basic point that I dont see, so I really need your help ..


Okay ..

so first off, we fire a stream of particle through a non-uniform B field and force acting on particle is

F = µ.grad(B)

we have splitting of beam due to different value of µ

µ = -Bohr magneton * ml (<<<<< please tell me if this true for ALL PARTICLES, not just electrons)

we have 2l+1 value of ml


my FIRST QUESTION IS:

1) the observation of physics shouldnt depend of the choose of coordinate right. Here we should have a fixed value of number of beam, i.e. we should have a fixed value of l. We can argue due to the conservation of angular momentum so l is fixed for any certain stream of particle.

But angular momentum is a quantity that depends on the choose of coordinate right? so we will have different value of l for different coordinate. This cant be true because then it will lead to different number of beam for different coordinate. PLEASE EXPLAIN



And then we have spin angular momentum coming up. s and ms. Keeping in mind the number of stream depends on the number of different value of magnetic moment. My second question is ..

2) What is the contribution of spin angular momentum to the magnetic moment?

I would think it is Bohr magneton * ms , and it would be additive to the contribution of orbital angular momentum. Is that true?

If so the number of beam would depnds on how many possible value of (ml + ms) right?


My third question is ....

in a postcard sent by Gerlach to Bohr, he describe he fired a source of silver atoms through a B field gradient, then the beam splitted into 2 beams.

3) Why only 2 beams? why the value of ml here = 0? also this is a multi electron system. how do you account for that?


Also when you fire a single electron through SG, due to 2 value of ms, there would be 2 beams coming out, but how about ml, does ml = 0, so does l = 0.

So particle moving straight through space has l = 0???? This is one of the main point i dont understand .


I know this is such a trivial question but please help

Thanks so much

Another very trivial question ..

I think it is due to I mix up between classical thinking and quantum thinking, but cant help.


We have L = h SQRT( l(l+1))
and Lz = h ml

we can see that L is never equal to Lz


So L will never be aligned in z direction??????


Another thing is ... we dont know Lx Ly simultaneously, so we cant never know when L is aligned in z direction or not, because then Lx Ly both = 0.


So we will never really know the direction of L, does that make sense?

Okay, here's the basic idea of the experiment (a very good account is in the first chapter of the Sakurai Modern Quantum Mechanics as an introduction to two-level quantum systems):

We have a furnace that heats up a sample substance (in the case of the original experiment, silver atoms). Those excited atoms, statistically, should have a completely random orientation of their angular momentum relative to the z-axis of the magnetic field. Therefore we expect a bell-shaped distribution of where the atoms end up after passing through the magnetic field. But the result is a few spikes with nothing in between, which indicates that the angular momentum must be quantized in the direction of the magnetic field, and iterated experiments tell us that things are overall quantized like that.

Since we are dealing with spin in this case, it doesn't behave like angular momentum in the classical sense. Spin is an intrinsic quantity, and isn't related to r x p in any real way. It turns out that it behaves a lot like angular momentum in terms of how the operators behave, and also that it couples with magnetic fields like how we would predict the angular momentum to create a magnetic dipole and couple with a magnetic field, so we call it "angular momentum."

I am sorry but that doesnt answer my question.
I understand about the concept of the experiment and all the concepts of angular momentum and intrinsic spin angular momentum, but my questions are different. Can you please go through question by question hightlighted blue above.

Can anyone please help. I know this may sound quite dumb to you but it will only take 5 mins.


Much appreciate.

(a) The stream will not have a uniformly zero angular momentum vector, but on average the ensemble will have zero angular momentum.

(b) The magnetic moment couples with the magnetic field, and that moment is a byproduct of the spin angular momentum, mostly. So then that particular value does not depend on reference frame. Spin 1/2 is spin 1/2 regardless of reference frame.

(c) The magnetic moment is proportional to the spin.

(d) The average particles passing through space has Lz zero, but this does not mean that they all have Lz = 0.

superlativity:

Welcome to sciforums.

In quantum mechanics, it is often true that the outcome of an experiment will depend on what you try to measure. In the SG experiment, the way the apparatus is set up determines the choice of the "z" direction of angular momentum. The "z" direction is specified by the orientation of the apparatus.

Not exactly. The spin and orbital angular momenta are vector quantities, and must be added as vectors. So, it is not a simple matter of adding the ml and the ms values to get the total angular momentum. In fact, the process of adding angular momenta in quantum mechanics is quite complicated, and is dictated by a set of mathematical objects called Clebsch-Gordan coeffecients.

The orbital and spin angular momenta of all the electrons in the Silver atom except for the outer shell electron cancel out to give zero net angular momentum. The outer electron has ml=0 and possible values of ms=+/- 1/2. Put the atoms through a SG apparatus and you get two results, corresponding to the 2 spin states.

That's right. If you picture L as a vector, it will never lie along the z axis.

In fact, one form of the uncertainty principle is:

(delta Lz)(delta phi) &gt;= hbar/2

where phi is the angle between L and the x axis, measured in the x-y plane.

Since we know Lz exactly, the uncertainty in phi is infinite, so we can't know anything about the x and y components of L. We can, however, know the magnitude of L.

Superlativity,
I am going to answer your questions, but you must follow along with the scenario uintil a thread is completed. The matter of "angular momentum" and "spin" is presently described in a mathematical ad hoc form. It does what it does, yet the model is certainly limited.

First of all the question of why silver havng only two possibilities? The answer given by James R focused on 'outer electron spin' being that electron providing various aspects supporting the qm notion. The 'moving electron' generating the "magnetic angular momentum' is the basic standard structure. Well, to me James R did not explain to my satisfaction of any underlying physical dynamics. Maintaining the scrutiny to models that expressly limit scrutiny is not a direction of research likely to return much in the way of physical dynamics in the "small world".

Try this.

A particle enters a magnetic field pointing with the field/gradient "up". Of 1,000,000 particles entering iit is 50/50 whether the particle state is 'up' or 'down'. It sure seems like a coin flip, but is this all the options open? Certainly not. Let us allow the experiment to carry us through.

Bare bones data says a particle will go up, or down, in a magnetic field/gradient. What can this mean, physiocally? The up down line is certainly set by the Stern_Gerlach field/gradient, but the up or down motion? This must be set by the particle, or the particle determines 'up' or 'down' along the Z-axis with respect to the lab frame. So the definiiton of the state of the particle is, xS, where S is the direction of the experimental apparatus along the z-axis of the lab and segment mfg. x is the +, or - direction of motion along S or the change in motion from the straightline trajectory. +S says, oriented with respect to the magnetic field/gradient, the particle goes 'up' where the field and gradient are both "up", where for the field direction, up is "south to north", floor to ceiling", pick a standard, and for the gradient, low energy density to high energy density of the magnetic field.The inverted peaked roof SG arrrangement will develop a gradient up from flat floor to peaked ceiling.

I think the next is critical. The spin element of the particle navigates as easily in an increasing energey density field as in a decreasing energy density field. It is magnetic, to be sure, and it has the identical physical quality of sailing against the wind, as with the wind, or without the wind, i.e straightline motion for the 010 or +/- state..

The current model has the state of the particle established in the "heat of the tungsten filament". Unfortunaltely this can not be determined directly by experiment. If we limit ourselves to observation, before imposing physical assumptions, we see the particle exhibits a 50/50 distribution of motion when entering the mfg. Limiting ourselves to data and events, the very first thing we notice is that the particle is affected by a magnetic field. A polarization state has been imposed on the particle. If you confine your thinking to boiling quantum states off tungsten filaments then go for it, I suppose.

There is a more drect approach; more supported by exparimental results. To assume the particle is generating the states as, for instance in time,

100 010 001 100 010 001 100 010 001 100 010 001

etc, The sequence is for a spin-1 particle having three possible states. This will not harm the generality of the thread, that instead of assuming a random generated sequence, we will simply go with the data and our model and assume the states are generated 123 123 123 and so on. Any other mathematicakl model is unnecessarily burdensome.Everyone knows what the answer is: 50/50 or 1/3, 1/3, 1/3.

The particle is moving along in a Y(000f) state meaning any appearing "1" indicates the current observed state; this is all it says. There are no physical assumptions at this point regarding the "unobserved states".

How can the particle magnetic polarization vector (mpv) always line up with the Z-axis of SG mag field/gradient direction? The simplest is with a magnetic monopole consfigured in a 'spherical' mode. A magnetic monopole' that will take the direction of the observed state when coming into contact with the mfg of the SG segment is the polarized state. If we assume the particle is virgin, unpolarized and hot off the tungstaen filament then our perfect model would hold and the current observed state would be manifest by the direction of motion. The symbol, +S tells us only which one oif the two or three state the particle is in, when measured.

The next logical step to take is determine what happens to the known state of a particle under various segment conditions?

Easy enough to do. Simply block all but one possible trajectory and the particle exiting is proved to be in one state only, that of the unblocked channel. Obstructiing the middle and lower channels in the spin-1 case, or the lower or - channel in the spin1/2 case, the exiting particle is a +S base state particle, with 100% confidence assuming a perfect gas and apparatus etc. In the spin-1 case 2/3 of the beam are lost to collisions with obstructions as are 1/2 of the particles lost to obstructed spin-1/2.

We have two options now open. Direct the +S particle into SG segments where one segment is unobstructed and the other wide open. We define a T segment as identical to an S segment but rotated around the axis of travel. Obstructions are not segment tyope diferentiating. Likewise, an S => T transition is defined as an alien-to-domestic transition (atd); an S-> S a domestic-to-domestic transition, (dtd). Some experimental gospel must now be imposed: any atd transition will always result in the S -> T transition upon entering the SG field/gradient region. If we write the transition as S ->T -> S this describes the transiton of an S particle through a wide open T segment. With one open channel only the particles exit the obstructed segment in the state they were polarized to when entering the magnetic field/gradient (mfg). This transiton then would read, S -> T + b -> T. Putting these two togethjer,
S -> T -> S
S -> b + T -> T
S -> b + S ->S

The third transition was added for instructional purposes. The blocking function (B) represneted by the "b" in the transition description is the only observed difference in the physics. The 2nd and 3rd expression give iidenticakl results as shown by application of the polarization rule. In the 2nd S is polarized to T in the atd transition. In the 3rd, the S state particle will always take the poalrized channel + her, and will not be subjected to state changes when transitioning through an exact replica of the particle's orgonal state. Block both middle and lower channels in the S segment and 100% of the +S particles will transition through, unscathed.

The 2nd sells us that the polarized state is interferrred with as the particle passes though the plane of the obstructions in the channels. The 1st expression says the particle is piolarized T until the trainsition T -> S occurs, which from he physical apparatus, can only be when the particle transitions fronm a mfg+ to a mfg0 or field to no field..

Furhter, there must be some elements of the +S state that are not expressed in the simple "+S" base statement. It will do no damage to add to the description the recognition of the application of forces through "unobserved" elements (we will refer to the elements as unobserved until such time they are observed). Unobserved at this point, 00[+S] added to the +S, would look like Y(1 00[+S]), where "1" referes to the "+" state, and the 00[+S] refer to those elements guaranteeing the reformation of the +S state. Spin-1 and spin-1/2 particles always survive the unobstructed and temporarily polarized state. What would a polarization process look like?

Y(000f) (f is the frequency of virgin observed state generation), any way,
(P)Y(1 00[+S]) -> Y( _ 00[+S] ---) -> Y(1 00[+S] 00[+T]), which describes an unambiguous hybrid, mixed state of he particle, which as we shall see is stable only in the presence of the mfg. The "_" are used to emphacize the noninstantaneous nature of physiocal proceses, and to emphacize the observation that the 00{+S] elements were unperturbed in the pol;arization and depolarization priocess. From the expression we see both entry and exit provideing exact replicas of each other, with one slight difference: Polarization goes from S -> T and depolarization goes from T -> S.

This state of the particle in the _ 00[+S] _ _ , nonlocal, or unobserved state, state is suffiecient to reform the original +S state of the aprticle. What are the elements, part of a spring loaded snapping back to exactly the +S state gizmo? S is respect to the particle's polarizing segment Z-axis which must be recovered in order to return to the observed +S state.

Magnetic compasses are directed north when released from any force perturbing thje equilibrium state of "north". The spin particles return north in tehe absence of an external field, which reads suscpiciously like some nonlocal force channels are at work or working together with mainatining the correct deviation from the +S direction fof the mpv..

In the second amd third expressions, the poalrized state sirvives, meaning that the obstructions are interferring with the 00{+S] and the 00[+T] when processing through obstructed segments; where are the obstructions? The obstrucitons are in the channels of the nonobserved elements defined at the event of polarization.
.
In the 2nd the obstructions perturbed equally the uunobserved elements of both possibe states, with, bingo, +T surviving or any possible xT, ( using +T does not harm generality). In the 3rd, the

(P)Y = Y(1 00[+S] 00[+S]) are perturbed, but any compoetition is detemined by the two headed coin. In the 2nd the 00[+T] always win over the competing 00[+S]), why?, because the 00[+T] retains the mpv direction for the T state when the particle state was determined by the force and drection of the mfg of the SG segment

The particles are inertial gyroscopic entities that have this complex mpv state generating system The pertubations are located many orders of magnitude away from the transitioning particle, which is observavbkly not affected by the obstructions. The exiting and entering states are the same in dtd transitions. In dtd transition we should assume that some hiccup, not amounting to a change of state, not observed state anyway , as the particle always takes the polarized channel.

I am dione fir now. If this is of any interest I will continue, If you have any question or discussion please feel free to say what you will. I will not be responding to claims such as, "'Quantum mechanisc' "does not allow this or that" replies.

This is not a dialogue on qm, this is not qm. If you respond to this quantum mechanically in opposition you are not just in the wrong pew you are in the wrong church. I say this from teh understanding that what I just said is the truth, the above is not qm; Anyione disputing this can of course do whay they will, but I am not in thsio for the debate.

I recognize parallels and similarities, but the foregoing was excluded from scrutuiny by the historical "fathers of quantum mechanics".

The previous state from whatever the mpv state was is completely erased from memory in transitions where only one channel is open. The exception bening thatg a polsrized particle will always take the channel of it origin when transitioning through the twin segment.

The polarized hybrid mixed state exists by virtue of he magnetic field alone. Remove the ecxternal field and the state either reformws in wide opnme transitions, or remains in the opolarixed state, noiw statble after excluding teh 00[+S] elements form the state structure

In the depolarization process Y(_ 00[+S] _ _) it is the unobserved elements 00[+S] that maintan the guaranatess for the reformation of the prepolarized state. The inclusion of the 00[+T] is to remanin consistent with the defintion of Y(000f), meaning only one observed lement is allowed at a time. a delta t~(1/2f). The state zipping through the channel is a, ]

(P)Y = Y(1 00[+S] 00[+T]), and the only other event is the depolarization process.

What happens? The +S paericle passing through the wide open segment exits that segment in the +S state, as if the segment wasn't there. The particles exiting the obstructed segment exit only in the state of polarization when entering the mfg of the SG segment.

geistkiesel

What value is tehre to state that the "ensemble" will have zero angular momentum? The particles behave exactly alike regarding their "anguilar momentum" which is not necessarily a significant parpameter regarding the spin attributes of he particle.

Again I think you neglect the obvious that the spin+/- 1/2 is always with respect to the particlular frame of reference where it is observedd. If we adopt a modle that the "magnetic polarization vector" always aligns with the manetic field of he SG segment then a natural conslusion is that the magnetuic polarization cvector is attached tot he particle as if secured by a 3 pi frictpnl;ess ubnivesal joint. The vector aligns wioth the field without torque applied and when the vector and field align then the particle will move in the polarized direction. This direction is always along the z-axis of the segment either up, or down for spin 1/2 particles. Taking spin-1 particles as an example, it should be clear that the magnetic properties of the magnetic polarization vector do not conform to an electron orbit model where the particle navigates as easily in an increasing energy density medium as in a decreasing energy density medium as well as in "straight line" motion. How does the standard model particle select a "south' or "north" pole, or a "neutral" pole? The answer is not found from analyses of standard model megnetic dynamics.

Maybe so, but what is being placed into a proportional equality? I look at "spin" as a simple direction indicator as seen in the deflection of straight line motion. By the time a magnetic moment has been observed all the activity regading magnetic moments is thoroughly divirced from the spin/polarization processes that give rise to the magnetic moment in the first instance. In the transition process there are two basic events: Polaization of the particle state when entering the SG segment and depolarization when exiting the segment - applied force and release of applied force each being sufficient to sustain the particle definiion as an inertial frame of reference where aplied force or relaease of fporce i ssufficient to alter the dynamics of he particle state. The assumption that the state is detemined in the heat of the tungsten filament is unproved and unprovable. If the tungsten filament were the generating force for the particle spin state, how does subsequent changes of state relate to the original polarization with in the heat if the filament? In other words in an obstruciton free SG segment the particle state upon exit from the SG segment, reverts to the state ovserved immedialtely prior to entering the magnetic field/gradient system of the current SG segment.

How can this be determined? If I understand you correctly you are assuming the the Lz for what orientation(?) being zero, all of them? I am not sure I understand the statement as made. For sure the spin states, for spin-1 particles all align with the segment z axis. There are no x or y components of spin that are observed. There are only x-axis components, for instance, used to define z axis spin stGeistkieselates from theory only. This is my understanding .

Geistkiesel

I would like to point out that, regardless of the orientation of your axes, you will still have a two-level spin-1/2 system if you rotate them again. So if I call up the z-axis at first, then rotate the SG device so that it's 45 degrees off from up, I'll still end up with a z'-axis that yields the same physical result.

If you have a perfectly random ensemble that is exactly equal, and you make measurements on, say, the z component of the spin angular momentum, you will find that, when you average the spins together after enough trials, that the average of the angular momentum is zero, but you would end up with an equal number of spin + and of spin -. That's basically one of the key results of the SG experiment aside from spin quantization.

I have a problem with the concept of the "average" of the spins of the beam. For sure, mathematically the paramteres measured that depend on the orientation of the magnetic spin vector the angular momentum averages to zeo, but what is the physical result here?

I undestand your statement that "the angular momenum average is zero, but ...an equal number of +1/2 and -1/2 spins", Why the stress on the word "but"? Isn't the average measured anguklar momentum the result of the equal distribution of observed +/- 1/2 spin states in the SG sement?

In other words, we aren't discussing equal angular momentum states in term sof Einstein-Podalsky-Rosen pespectives and "twin" photons. It is only apparent that the very natural +/- spin states of the particle are determined wrt to the SG segment z-axis, but, the orientation of spin is determined by the particle after the polarization along the z-axis is established. Simply said the electron can move as easily along the + as well as the - direction even though the energy density of the field gradient system is 'small to large' in one case and 'large to small' in the other. Simple minded + and - energy states do not adequately express the physics of the induced spin motion.

Geistkiesel

What exactly as "the energy density of the field gradient system" I have no idea what that means?

And what does the EPR paradox have to do with anything that was stated?

PhysMachine - This is as complete and spartan a description as I am able to produce. The transition results that are experimentally determined can be verified from Feinman's "Lectures on Physics", Vol. III, Chanpter 5. I agree with the bare bones experimenatl results described by Feinman but am polarized to an unambiguosly differerent interpretation of the physics.

The delay in answering your post could not be avoided. Also, in another post I refered to the EPR experiment meaning that no entanglement issues are apparent and therefore the discovery that the sum of all transitioning particles in the beam's angular momentum is zero is without informative value.

S and T segments differ only in the angle of relative rotation around the y-axis (the unpolarized direction of particle moption) between the field/gradient directions of the two segments.The spin state of a spin-1 particle may be predicted more accurately using Stern-Gerlach rules of transition than any current mathematical model.

By “spin state” I mean

1. the current (or future) direction of the spin-1 particle motion off the other wise unperturbed y-axis vector, and
2. where the particle is perturbed by the magnetic field/gradient of the SG segment only.

Any other model, such as using “superposition of states” may appear competitive, or contradictory even, but a simple analysis will prove this fear in error. If I predict the particle will definitely and use a “+” or upper channel of a consecutive SG segment transition, I mean that the particle left the previous SG segment in the +S state, unambiguously.

The previous SG segment can be either

1. an S segment where the “+/-“ and “-“ channels are completely blocked, or
2. a T or S segment where all channels are unobstructed, or “wide open”.

The basic and complete list of state transition rules is as follows:

1. All spin-1 particles entering a SG field/gradient volume are polarized to one of three possible direction changes:
   1. plus, or “+’ or “up”, or
   2. plus/minus, or, “+/-“ or ‘horizontal, or
   3. minus, or “-“, or down.
2. The magnetic polarization vector, ("spin state vector") may be considered as spherical until perturbed by the field/gradient forces which distor the "perfect" spherical magnetic monopole.
3. Spin states are not generated in the heat of the tungsten filament that produced the spin-1 particles for observatioopn. The internal generation of spin states is an intrinsic attribute of spin-1 particles. These state may be described as ocurring as 100 010 001 100 010 001 etc where it is understood that only a "1" describes an observed, or observable state. The ")" described unobserved states. There is no need to contruct a random "three spin state generator" when a hard wired ieterative meaning 123 123 123 etc generation mode will return identical results.
4. All S particles are polarized to the specific host segment magnetic field/gradient arrangement when first sensing the effective field/gradient force of the host T SG segment (or S segment).. These S to T transitions are “alien-to-domestic” transition. All S-to-S transitions are domestic-to-domestic transitions and produce no observable inner state changes in the particle.
5. All spin-1 particles exit unobstructed SG segments during de-polarization with the pre-polarized spin state. A particle in a “-S” spin state transitions through a T segment as either a “+T”, “+/- T” or a “-T” T state and reforms to the “-S “ state upon exit from the T segment.
6. Two channel obstructions in a T segment will produce the final spin-state defined by the unobstructed channel. For example with the “+” and “-“ channels obstructed; only the “+/-” spin state is produced.
7. A +S particle transitioning through an obstructed “+T” and “+/-T” segment will exit the T segment in a “-T” state and upon entering a “+S” SG segment (identical to the first that produced the initial “+S” state) will exit the final S segment in the +S spin state, where the +S state here is not determined in any way by the original “+S” particle state. The second segment, a doubly blocked T segment, erases all memory of the original “+S” spin state.
8. Without excepetion all magnetic polarization vectors of spin-1 particles adopt the host segment magnetic field/gradient directions upon entry into the host segment.
9. Doubly obstructed T segments will completely obstruct 2/3 of the particle beam in alien-to-domestic transitions. Unobstructed alien-to-domestic and domestic-to-domestic, or obstructed domestic-to-domestic transitions where the unobstructed channel is identical in description to the transitioning particle spin state produce no beam losses in transitioning particle count.

James R, I do not dispute your understanding of the QM explanation of angular momentum, but I do have a question(s).

If we focus on spin-1 particles (three possible spin states +, +/- an - , I use +/- instead of 0) there seems to be a void in explaining the nature if the magnetic state of the particles. Particles that are seen to go up along the Z a-axis, say +S, do so with the same ease that a -S particle moves down. As the field/gradient of the SG segment can be described as a gradient of magnetic field energy a particle navigating in an increasing energy field is equivalent, or nearly so, to the particle navigating in the decreasing energy field. Simple +,- factors for the two aren't satisfactory descriptions as analogos to electron charges. The particles themselves are identical and exhibit motion differences only. I havae a feeling that magnetic monopoles are in abundance, though lost in the classical description of "orbiting electrons".

I am familiar with the orbiting electron model for the creation of magnetic fields, but this model is wanting for the lack of strong experimental proof and it seems contrived. What is your take on this? AS all angular momentum is seen in the z-axis, +, +/- and - , how does the QM world justify x components of z=axis angualr momentum never having discovered, observed or measured such off axis values?
Geistkiesel

What do you mean by "beam splitting" Superlativity?
And what form do the different values of "u" take that explains the different motion states to you?

Geistkiesel

EPR had nothing to do with th ediscussion which is why I commented that the total angular momentum averaged at zero has no physical meaning and certainly entanglement was not an issue here.
Here is what we are basically working with.

The figurfe shows the inhomogeneous field and the homogeneous field. Also a schematic of the SG segments is shown where the T segemnt follows after an unobstructed S segment. The diagrams showing how the transitions go as per experimental results shows a simple way to by pass qm mathematical predictions (which are of statitical accuracy only) and have the unambiguous current energy state of transitioning particles determined by inspection. The obstructed channels have an operational effect in the transition of the particles. Apparently the unused states fill a function while in the SG field environment as nonlocal force centers and as essential to the reformation of the original state in the unbstructed case,. These nonlocal attributes are force centers that are essential in the particle's magnetic spin vector's reorientation back to the originally polarized state, while in the absence of any field. The particle is temporarily polarized to some T state and only when exiting he force field is the reforamtion back to the input state accomplished.

The spin-1 particle is essentially an inertial platform that finds its own north in the absence of any observed force field.

Geistkiesel