# Understanding the Wave Function and Excited States

A wave function is an infinite dimensional vector space, how can it "live" in $\mathbb{R}^3$?

Given the equation that is built like: $$\Psi (x,t) = \sum ^{\infty} _{n=1} c_n \psi _n (x) e^{-i E_n t / \hbar}$$

How does one "excite" a quantum particle in the lab? The excited states simply give a different probability distribution for the particle?

For the infinite square well, the wave function in excited state $n$ has exactly $n$ bases, all of which are non zero? If a particle is in its ground state, does that mean that all other bases must be zero? making the wave function dependent only on one basis?

• As "Programmer" mentions, I think you're confusing the Hilbert space of possible wavefunctions with the real-space representation of a particular element. Also, you write "A wave function is an infinite dimensional vector space", whereas I think you meant to write "A wave function is an element of an infinite dimensional vector space". Oct 25, 2013 at 3:24

Let's go step by step:

1) A particle (or a system in general) can be found in a given state $|\psi\rangle$, which mathematically is defined as a vector of an Hilbert state $\mathcal{H}$. This is to ensure that the concept of scalar product (necessary to define ortonormal basis) and norm (necessary to define the probability amplitude) are well defined. When you then decide to expand your wavefunction on the position basis you get something like: $$|\psi\rangle = \mathbb{1}|\psi\rangle = \int dr |r\rangle\langle r|\psi\rangle = \int dr\, \Psi(r)|r\rangle$$

where I defined $\Psi(r): \mathbb{R}^3\rightarrow\mathbb{C}$ as the wavefunction of the state $|\psi\rangle$ represented at the position $r$. If you want your wavefunction at position $x$, all you have to do is multiply by $\langle x|$ the above expression: $$\langle x|\psi\rangle = \int dr \,\Psi(r)\langle x|r\rangle = \int dr \,\Psi(r)\delta(x-r) = \Psi(x)$$

and there you go.

2) Suppose now that you want to perform a measurement to characterize your system. Mathematically an operator $\hat O$ is associated to the measurement process, which you suppose has a complete orthonormal set of eigenvalues $\{|\psi_n\rangle \}_{n=0}^\infty$. You can then expand the wavefunction associated to your system as in the formula you provided. Now, since you choose a set of basis functions, what specifies the wavefunction is the coefficients $c_n$ of the expansion. Therefore, if you perturbe the system, your wavefunction will have another set of coefficients $c'_n$ onto the same basis. Since the probability amplitude is defined as the square modulus of the wavefunction, you'll get (exercise!) $$|\Psi(x,t)|^2 = \sum_n |c_n|^2$$ therefore a different set of coefficients will give you a different probability distribution.

3) If you are in the ground state of the system, the result of the measurement will be $$\hat O|\psi\rangle = |\psi_0\rangle$$ every time you perform the measurement (see the Von Neumann's postulate). This means that in the coefficient expansion only $|c_0|=1$, while or the other coefficients are zero. It's really wrong to say that the basis terms must be zero, the basis is just the basis and it's associated to the measurement operator, it's the coefficients that matter for the wavefunction. Also "making the wavefunction depend on one basis" doesn't make much sense. Of course the representation of the wavefunction depends on the basis of choice, but you choose the basis with respect to the measurement you want to perform on it, i.e. with respect to the operator you're considering.

• Ok, now you explain the down-vote to me. Come out and explain step by step what's wrong! Oct 31, 2013 at 8:18

A wave function is an infinite dimensional vector space, how can it "live" in ℝ3?

The wave function does not "live" in $\mathbb{R}^3$ but in Hilbert space $\mathbb{C}^{\infty}$.The probability amplitude of ﬁnding the particle at position $x$ when in a known state $|n \rangle$ evaluated as $\langle x|n \rangle$ could describe the particles position in $\mathbb{R}^3$. Infinite-dimensional spaces provide headaches for physicists and employment for mathematicians.

How does one "excite" a quantum particle in the lab?

Using photons (electric fields + magnetic fields). They can change the kinetic energy, spin and angular momentum or in short, "excite" the particle.

The excited states simply give a different probability distribution for the particle?

It changes the wave function from a state $\langle i\rangle$ to a different state $\langle f \rangle$and therefore changes the probability distribution.

If a particle is in its ground state, does that mean that all other bases must be zero?

Yes. If you know it is in state $\langle n\rangle$, then it has energy $E_n$. Unless you "excite" the particle, it will remain in state $\langle n \rangle$. End of story.

making the wave function dependent only on one basis?

The state can be described using one basis. The wave function must describe all the possible states, which are infinite.

• Sorry, but the last two sentences don't make much sense. It's misleading to say that the measurement excites the particle. I think you refer to the Von Neumann postulate, that a measurment on a state brings it in an eigenstate of the operator associated with the measurement. You could have some other external perturbation which now excite the system, but if you repeat the measurment you'll get another (different) eigenstate of the measurement operator. Oct 25, 2013 at 7:04
• Then "The wavefunction must describe all the possible states" it's just wrong. The wavefunction simply describe the state of the system, I think you meant that in the expansion of the wf on a orthonormal set you must include all the terms of the given set, which are infinite. Oct 25, 2013 at 7:06
• As far as "live in" should be used at all, I'd not say "the wave function lives in $\mathbb R^3$" is wrong talk. On the other hand, I'd much more have a problem with "does that mean that all other bases must be zero?". I also don't understand your last sentence: "The state can be described using one basis. The wave function must describe all the possible states, which are infinite." Oct 25, 2013 at 7:07
• I agree with you, as you can see, I stated the same comment ;-) Oct 25, 2013 at 7:26
• @Programmer Mattia is correct here, if you solve the equation you may get all possible eigenstates, but a given solution is a single state. Superpositions are states too, even if not eigenstates. Nov 7, 2013 at 14:37