Complex conjugate of a wavefunction

  • #1
Let ψ be a wavefunction describes the quantum state of a particle at any (x,t), What does ψ* i.e, the complex conjugate of a wavefunction means? I only know probability of finding a particle is given by ∫|ψ|^2 dx= ∫ ψ*ψ dx But what does ψ*ψ really means? I started learning QM with Griffiths without enough mathematical background, Help me.
 

Answers and Replies

  • #2
470
58
As you said, supposing that you have a wave function ψ(x) describing your particle in space, then [itex]\int_a^b \psi^*(x)\psi(x)dx[/itex] is the probability of finding your particle between the points a and b.

It follows that [itex]\psi^*(x)\psi(x)[/itex] is the so called probability density. It is nothing but the function that, when integrated, gives you the probability itself.
 
  • Like
Likes 1 person
  • #3
19
4
If you measure the position of the particle, at time t, ψ*ψ(x,t) represents the likelihood that you will find it to be at position x... or more accurately within between x and x+dx.

As for what ψ* means, it's the complex conjugate of the wavefunction... but of course you already know that. If you're looking for a nice intuitive way to interpret the physical meaning of the phase of a wavefunction, then I'm afraid I don't think that there is one.
 
  • Like
Likes 1 person
  • #4
470
58
As for what ψ* means, it's the complex conjugate of the wavefunction... but of course you already know that. If you're looking for a nice intuitive way to interpret the physical meaning of the phase of a wavefunction, then I'm afraid I don't think that there is one.
Exactly, this is just how the mathematical formalism of Quantum Mechanics works.
 
  • Like
Likes 1 person
  • #5
ChrisVer
Gold Member
3,360
456
you only need [itex]\psi^{*}[/itex] because you need to have a real thing to integrate over to get probability.... [itex] \psi[/itex] alone is complex, and also it's arbitrary up to an overall phase.... that means that either you take [itex]\psi[/itex] or [itex] e^{i \theta} \psi[/itex] they both are solutions of the same shrodinger equation. As such you must make it real, or you'd have arbitrary probabilities depending on a phase...
One quantity that is for sure real is [itex] \psi^{*} \psi [/itex] and nothing is "lost" from [itex]\psi[/itex]'s initial parameters (if for example you write [itex]\psi=a+ib[/itex] the result will contain both a and b...

Another example in which you can see that, is when you have [itex]\psi[/itex] real, where you don't need the complex conjugate and just take the square....

In fact the complex conjugate's "meaning" (mathematical) becomes clearer when you start working on the states in vector spaces... The conjugate appears as a relation between the bras and kets, and thus between the vector and its dual space.
 
  • #6
9,504
2,590
Let ψ be a wavefunction describes the quantum state of a particle at any (x,t), What does ψ* i.e, the complex conjugate of a wavefunction means?
Its got to do with the bra-ket notation:
http://en.wikipedia.org/wiki/Bra–ket_notation

The wavefunction is looked at as the Ket - its conjugate as the Bra - apply a Bra to a Ket and square it to get probabilities - that's the Born rule.

At a deeper level states are not actually Bra's or Kets - they are really operators - what are normally referred to as states have a special name called pure.

At an even deeper level again it is seen states are really what's required so we can work out probabilities of the outcomes of observables - this is the important Gleason's Theorem (the key assumption is what is known as non contextuality you may have heard about):
http://en.wikipedia.org/wiki/Gleason's_theorem

It's a devil of a thing to prove however so its not usually talked about much at the beginning level despite its importance.

There is a variant that is much easier to prove and some people use it as the foundation of QM eg:
http://arxiv.org/pdf/quant-ph/0205039v1.pdf

Anyway that will take you to a pretty advanced level. For now what I would do is get a hold of Ballentine:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20&tag=pfamazon01-20

Its mathematically quite a bit more advanced than Griffiths. But hold onto your nerves and give the first 3 chapters a read. You wont likely understand too much of the detail, but, hopefully, fingers crossed, you will get the gist so that things are a lot clearer, and Griffiths will be easier. Once you have finished Griffiths you can move onto Ballentine.

Thanks
Bill
 
Last edited by a moderator:
  • #7
UltrafastPED
Science Advisor
Gold Member
1,912
216
Let ψ be a wavefunction describes the quantum state of a particle at any (x,t), What does ψ* i.e, the complex conjugate of a wavefunction means? I only know probability of finding a particle is given by ∫|ψ|^2 dx= ∫ ψ*ψ dx But what does ψ*ψ really means? I started learning QM with Griffiths without enough mathematical background, Help me.
ψ is an element of a Hilbert space; ψ* is an element of it's dual space.

The Dirac bra-ket notation makes use of this: |ψ> is an element of the Hilbert space, and <ψ| is it's dual.

Following the general conventions for vector spaces with an inner product, we have <ψ|ψ> is the inner product, and it generates a real number.
 

Related Threads on Complex conjugate of a wavefunction

  • Last Post
Replies
6
Views
2K
Replies
2
Views
634
  • Last Post
2
Replies
41
Views
12K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
4
Views
7K
Replies
3
Views
13K
  • Last Post
Replies
2
Views
546
Replies
3
Views
1K
Replies
1
Views
3K
Replies
4
Views
2K
Top