# Difference between wave and matrix QM

1. Aug 18, 2010

### snoopies622

I'm still enjoying Daniel T. Gillespie's "A Quantum Mechanics Primer". It includes six postulates of quantum mechanics which are very similar to sets I've found in other texts, as well as by internet-searching "quantum mechanics postulates":

1. Identifying the state of a system with a vector in Hilbert space

2. identifying observables with Hermitian operators

3. the formula for predicting the results of a measurement

4. the effect of a measurement of the state vector itself

5. the time evolution of the state vector (the Schrodinger equation)

6. specifically defining the position and momentum operators, and the way they can be used in combination to form new operators.

I was wondering, is there any difference between these postulates and the Heisenberg formulation of QM (or what might still be called "matrix mechanics") other than substituting his formula of the time evolution of an operator for the Schrodinger formula of the time evolution of a state?

2. Aug 18, 2010

### vanhees71

I don't know this book, but the postulates are somehow a bit unprecise.

First of all the (pure) states of a quantum system are not represented by the vectors of a Hilbert space but by equivalence classes of such vectors, namely the rays in Hilbert space. That boils down to phase invariance and is crucial for quantum theory to describe the real world (e.g., there wouldn't be non-relativistic Galileo invariant QM if you postulate that the states are represented strictly by Hilbert-space vectors; also that there exist particles with spin 1/2, i.e., that the rotation group SO(3) can be substituted by its covering group, SU(2)) etc.).

Further the observables must be represented not only by Hermitean operators, but the operators must be even essentially self-adjoint.

Postulate 4 is too unspecific. What is written there? If a collapse is postulated, it's at least problematic. For sure it's not necessary of the successful application of quantum theory to the real world.

The time evolution of the states is given by the Schrödinger equation only in the Schrödinger picture of time evolution. Other pictures (like the Heisenberg picture or the most general Dirac picture, of which the interaction picture is an important example) are important in practice.

3. Aug 19, 2010

### snoopies622

Thanks for all the information!

Yes, sorry about that - I wrote them imprecisely. I didn't want to bother entering all the equations and I thought everyone would know what I was talking about. Maybe instead I should have asked, what is the difference between the Schrodinger formulation and the Hiesenberg formulation? As far as I know, the only difference is that the first one uses

$$i \hbar \frac {d \psi}{dt} = \hat {H} \psi$$

and the second uses

$$[\hat {A} , \hat {H}]= i \hbar \frac {d \hat {A} } {dt} \vspace {5 mm}.$$

Last edited: Aug 19, 2010
4. Aug 19, 2010

### Fredrik

Staff Emeritus
What you're describing is just the difference between the Schrödinger picture and Heisenberg picture of the Hilbert space version of QM, which was invented by von Neumann. I'm not familiar with Schrödinger's and Heisenberg's original work, but I suspect that their theories are just von Neumann's theory for specific Hilbert spaces. In Schrödinger's case, it would be $L^2(\mathbb R^3)$, and in Heisenberg's, any finite-dimensional complex Hilbert space would do (so that vectors and operators can be represented by matrices).

(I could be completely wrong about what Heisenberg's theory was).

5. Aug 19, 2010

### vanhees71

That's not correct. Heisenberg and Schrödinger picture of time evolution are only two representations of the same quantum theory. Also the Hilbert spaces are the same. There's up to equivalence only one separable complex Hilbert space, and that's the one we are talking about in non-relativistic quantum theory for a fixed number of particles.

Let's take the quantum theory of one particle with a time-independent Hamiltonian which makes the formulae a bit easier to write than in the general case. In the Schrödinger picture by definition the state kets carry the full time dependence, i.e.,

$$|\psi,t \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi,0 \rangle$$

and (not explicitly time-dependent) observables are constant in time

$$\hat{O}(t)=\hat{O}(0)=\mathrm{const}.$$

The probability amplitudes (wave functions) are the (generalized) scalar product of a (generalized) eigenvector with the state ket:

$$\psi(t,o)=\langle o | \psi,t \rangle = \langle{o}| \exp(-\mathrm{i} \hat{H} t|\psi,0 \rangle.$$

In the Heisenberg picture, the operators representing observables carry the total time dependence, i.e., one has

$$\hat{O}_{\mathrm{H}}(t)=\exp(\mathrm{i} \op{H} t) \hat{O}(0) \exp(-\mathrm{i}$$ \hat{H} t),$$which implies that the (generalized) eigenvectors of these operators become time dependent via $$|o,t \rangle_{\mathrm{H}}=\exp(\mathrm{i} \hat{H} t) |o,0 \rangle_{\mathrm{H}}.$$ The state kets in the Heisenberg picture are constant in time: $$|\psi,t \rangle_{\mathrm{H}}=|\psi,0 \rangle_{\mathrm{H}}=\mathrm{const}.$$ The wave function at time $$t$$ now reads $$\psi(o,t)=_{\mathrm{H}} \langle o,t|\psi,t \rangle_{\mathrm{H}} = _{\mathrm{H}} \langle o,0|\exp(-\mathrm{i} \hat{H} t)|\psi,0 \rangle_{\mathrm{H}},$$ where I used the self-adjointness of $$\hat{H}$$. Since the Hamiltonian is the same in both pictures and since at $$t=0$$ by definition the state kets and the (generalized) eigenvectors are the same in both pictures, this result is the same in both pictures as it must be for an observable quantity as the probility to measure a certain value of the observable $$O$$, which is given by $$|\psi(o,t)|^2$$. Another aspect of the formulation of QT by Heisenberg/Born/Jordan and Schrödinger is the representation. Heisenberg used the harmonic oscillator as one of the most simple examples to propose his theory, and he used the discrete set of energy-eigenvectors as basis. In such a basis the wave functions are represented by an infinite dimensional'' column vector with components $$\psi_n(t)=\exp(-\mathrm{i} E_n t) \langle E_n|\psi,t \rangle$$ (which is of course again independent of the picture used; here I used the notation in the Schrödinger picture for convenience). The Hilbert space is thus realized as the Hilbert space of square summable series, $$\ell^2$$, with the components, $$\psi_n(t)$$ as summands. Schrödinger instead used the position representation, i.e., $$\psi(x,t)=\langle x|\psi,t \rangle.$$ which leads to the realization of the quantum theoretical Hilbert space as the space of square-integrable functions, $$\mathrm{L}^2$$. Of course, as is very clear in Dirac's representation-independent bra-ket notation, both realizations of quantum theory are equivalent. 6. Aug 19, 2010 ### vanhees71 The Heisenberg and Schrödinger picture of time evolution are only two representations of the same quantum theory. Also the Hilbert spaces are the same. There's up to equivalence only one separable complex Hilbert space, and that's the one we are talking about in non-relativistic quantum theory for a fixed number of particles. Let's take the quantum theory of one particle with a time-independent Hamiltonian which makes the formulae a bit easier to write than in the general case. In the Schrödinger picture by definition the state kets carry the full time dependence, i.e., $$|\psi,t \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi,0 \rangle$$ and (not explicitly time-dependent) observables are constant in time $$\hat{O}(t)=\hat{O}(0)=\mathrm{const}.$$ The probability amplitudes (wave functions) are the (generalized) scalar product of a (generalized) eigenvector with the state ket: $$\psi(t,o)=\langle o | \psi,t \rangle = \langle{o}| \exp(-\mathrm{i} \hat{H} t)|\psi,0 \rangle.$$ In the Heisenberg picture, the operators representing observables carry the total time dependence, i.e., one has $$\hat{O}_{\mathrm{H}}(t)=\exp(\mathrm{i} \hat{H} t) \hat{O}(0) \exp(-\mathrm{i}$$ \hat{H} t),$$

which implies that the (generalized) eigenvectors of these operators become time dependent via

$$|o,t \rangle_{\mathrm{H}}=\exp(\mathrm{i} \hat{H} t) |o,0 \rangle_{\mathrm{H}}.$$

The state kets in the Heisenberg picture are constant in time:

$$|\psi,t \rangle_{\mathrm{H}}=|\psi,0 \rangle_{\mathrm{H}}=\mathrm{const}.$$

The wave function at time $$t$$ now reads

$$\psi(o,t)=_{\mathrm{H}} \langle o,t|\psi,t \rangle_{\mathrm{H}} = _{\mathrm{H}} \langle o,0|\exp(-\mathrm{i} \hat{H} t)|\psi,0 \rangle_{\mathrm{H}},$$

where I used the self-adjointness of $$\hat{H}$$. Since the Hamiltonian is the same in both pictures and since at $$t=0$$ by definition the state kets and the (generalized) eigenvectors are the same in both pictures, this result is the same in both pictures as it must be for an observable quantity as the probility to measure a certain value of the observable $$O$$, which is given by $$|\psi(o,t)|^2$$.

Another aspect of the formulation of QT by Heisenberg/Born/Jordan and Schrödinger is the representation. Heisenberg used the harmonic oscillator as one of the most simple examples to propose his theory, and he used the discrete set of energy-eigenvectors as basis. In such a basis the wave functions are represented by an "infinite dimensional'' column vector with components

$$\psi_n(t)=\exp(-\mathrm{i} E_n t) \langle E_n|\psi,t \rangle$$

(which is of course again independent of the picture used; here I used the notation in the Schrödinger picture for convenience). The Hilbert space is thus realized as the Hilbert space of square summable series, $$\ell^2$$, with the components, $$\psi_n(t)$$ as summands.

Schrödinger instead used the position representation, i.e.,

$$\psi(x,t)=\langle x|\psi,t \rangle.$$

which leads to the realization of the quantum-theoretical Hilbert space as the space of square-integrable functions, $$\mathrm{L}^2$$.

Of course, as is very clear in Dirac's representation-independent bra-ket notation, both realizations of quantum theory are equivalent.

7. Aug 19, 2010

### Fredrik

Staff Emeritus
I appreciate the effort, but I know what the Schrödinger and Heisenberg pictures are. You misread my post. My point was that the difference between the Schrödinger and Heisenberg pictures in modern QM, is a completely different issue than the difference between Schrödinger's and Heisenberg's theories.

8. Aug 19, 2010

### vanhees71

Then I misunderstood you. I hope, it is also clear to you now that Schrödinger's and Heisenberg's theories are one and the same quantum theory and that the Hilbert space is in both cases the same either.

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