10 Identical bosons

As you will remember, every par­ticle is either a boson or a fermion. Let us use the symbol |a> for the state of a boson, and let us keep firmly in mind that a “state” in quantum mechanics is a prob­a­bility algo­rithm; it serves to assign prob­a­bil­i­ties to pos­sible out­comes of mea­sure­ments — in this case any mea­sure­ment to which the boson may be subjected.

As said, quantum states (and thus the pos­sible out­comes of mea­sure­ments that may be made) are deter­mined by the actual out­comes of mea­sure­ments that have been made. The symbol |a> accord­ingly plays a double role: it rep­re­sents (i) the out­come of a mea­sure­ment and (ii) the algo­rithm to be used for assigning prob­a­bil­i­ties to the pos­sible out­comes of whichever mea­sure­ment is made next.

If we want to know the prob­a­bility with which a boson ini­tially described by |a> is later found in the state |b>, we need to know the ampli­tude asso­ci­ated with this pos­si­bility. For this we shall use the symbol <b|a>.

Now sup­pose that a boson has been found “in” the state |a>: it is cor­rectly and com­pletely described in terms of the prob­a­bil­i­ties that |a> serves to assign. Also sup­pose that another boson has been found “in” the state |b>: it is cor­rectly and com­pletely described in terms of the prob­a­bil­i­ties that |b> serves to assign. (We shall fur­ther assume that |a> and |b> are pos­sible out­comes of the same mea­sure­ment so that, tech­ni­cally speaking, they are orthog­onal.) What symbol shall we use for the state of the com­posite system made up of the two bosons?

If we use some­thing like |a,b>, we intro­duce into our nota­tion a dis­tinc­tion that does not cor­re­spond to any­thing in the actual world. In addi­tion to the phys­i­cally war­ranted dis­tinc­tion between the boson described by |a> and the boson described by |b>, there is then the phys­i­cally unwar­ranted dif­fer­ence between the “left” boson and the “right” boson. To elim­i­nate this phys­i­cally unwar­ranted dif­fer­ence, we use the sym­metric symbol

(|a,b> + |b,a>)/√2.

The divi­sion by √2 ensures that this two-​​boson state is nor­mal­ized: the prob­a­bil­i­ties it assigns to the pos­sible out­comes of any mea­sure­ment to which the two-​​boson system may be sub­jected add up to 1. Observe that this expres­sion remains unchanged if we exchange |a> and |b>.

If, on the other hand, the dis­tinc­tion between the “left” boson and the “right” boson cor­re­sponds to some­thing in the actual world — if, for instance, the “left” and the “right” boson are of dif­ferent types so that “left” and “right” rep­re­sent the respec­tive types to which they belong — then the symbol |a,b> may be used, and we can write

<c,d|a,b> = <c|a> <d|b>

for the ampli­tude asso­ci­ated with the tran­si­tion from |a,b> to |c,d>. In this case, how­ever, the bosons are not com­pletely described by the indi­vidual states |a> and |b>, inas­much as these con­tain no infor­ma­tion about the par­ticle species to which each boson belongs.

In case the bosons are com­pletely described by |a> and |b>, the cor­re­sponding tran­si­tion ampli­tude is

(1/√2)(<c,d| + <d,c|) × (1/√2)(|a,b> + |b,a>).

The manip­u­la­tion of this expres­sion is straightforward:

(<c,d|a,b> + <c,d|b,a> + <d,c|a,b> + <d,c|b,a>)/2

= (<c|a> <d|b> + <c|b> <d|a> + <d|a> <c|b> + <d|b> <c|a>)/2

= <c|a> <d|b> + <d|a> <c|b>.

The tran­si­tion prob­a­bility is thus given by

p = |<c|a> <d|b> + <d|a> <c|b>|2.

This should remind you of the prob­a­bility p = |A(N→E,S→W) + A(N→W,S→E)|2 we pre­vi­ously obtained. Just put <E|N><W|S> and <W|N><E|S> in place of A(N→E,S→W) and A(N→W,S→E), respectively.

Sup­pose, then, that an ini­tial mea­sure­ment indi­cated the pres­ence of two indis­tin­guish­able bosons described, respec­tively, by |a> and |b>, and that the next (rel­e­vant) thing that can be deduced from an actual event or state of affairs is the pres­ence of two bosons described, respec­tively, by |c> and |d>. Is the boson that was described by |a> the same as the boson that is now described by |c> (in which case the boson that was described by |b> is the same as the boson that is now described by |d>)? Or is the boson ini­tially described by |a> the same as the boson now described by |d> (in which case the boson ini­tially described by |b> is the same as the boson now described by |c>)? Which boson in the ini­tial state of the two-​​boson system is iden­tical with which boson in the final state?

Do these ques­tions have answers?

Once again they don’t, but this time the reason is not simply that, if they had, the tran­si­tion prob­a­bility would be |<c|a><d|b>|2 + |<d|a><c|b>|2 rather than |<c|a><d|b> + <d|a><c|b>|2. Instead, we have put our finger on the reason both why these ques­tions have no answers and why the tran­si­tion prob­a­bil­i­ties come out the way they do.

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